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'ALGEBRA 


FOR 


SECONDARY    SCHOOLS 


BY 


WEBSTER   WELLS,   S.B. 

PROFESSOR  OF  MATHEMATICS   IN  THE  MASSACHUSETTS 
INSTITUTE  OF  TECHNOLOGY 


BOSTON^,   U.S.A. 

D.  C.   HEATH  &  CO.,  PUBLISHERS 

1906 


Copyright,  1906, 
By  WEBSTER  WELLS. 

All  rights  reserved. 


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PREFACE 

The  present  work  is  intended  to  meet  the  needs  of  High 
Schools  and  Academies  of  the  highest  grade.  While  in  the 
main  similar  to  the  author's  "  Essentials  of  Algebraj"  many- 
additional  topics  have  been  introduced,  and  improvements 
made;  attention  is  especially  invited  to  the  following: 

1.  The  product  by  inspection  of  two  binomials  of  the  form 
mx  +  n  and  px  -\-  q  (%  100). 

2.  In  the  chapter  on  Factoring  will  be  found  the  factoring 
of  expressions  of  the  forms  x^  +  ax^y^  -f  y*  and  aar'  +  6a;  +  c, 
when  the  factors  do  not  involve  surds  (§§  115,  117). 

These  forms  are  considered  later  in  §§  298  and  300. 
The  solution  of  equations  by  factoring  is  also  taken  up  in 
this  chapter. 

In  §  107  will  be  found  many  new  varieties  of  examples. 

3.  In  §  176  will  be  found  a  set  of  problems  in  which  the 
solutions  are  negative,  fractional,  or  zero. 

4.  In  the  chapter  on  Evolution  will  be  found  the  square 
root  by  inspection  of  polynomials  of  the  form 

a^  4-  62  _^  c2  +  2ab  +  2ac  +  26c, 

and  the  cube  root  by  inspection  of  polynomials  of  the  form 
a'  +  S a'b  +  Sab'-h  b'  (§§  212,  223). 

The  development  of  the  rules  for  the  square  and  cube  root 
of  polynomials  and  arithmetical  numbers  leaves  nothing  to  be 

22l7b8 


iv  PREFACE 

desired  from   a  theoretical  point  of  view.     (See  §§  213,  214, 
217,  224,  225,  228.) 

5.  In  the  solution  of  quadratic  equations  by  formula  (§  289), 
the  equation  is  in  the  form  ax^  -\-  bx  -{-  c  =  0. 

6.  In  all  the  theoretical  work  in  Chapter  XXI,  the  quadratic 
equation  is  in  the  form  ax^  -\-  bx  -{-  c  =  0. 

7.  In  the  chapter  on  Ratio  and  Proportion,  in  several  of  the 
demonstrations  of  theorems,  fractions  are  used  in  place  of 
ratio  symbols. 

8.  In  §§  386  and  387  will  be  found  the  same  proof  of  the 
Binomial  Theorem  for  Positive  Integral  Exponents  as  is  given 
in  the  "  Essentials  of  Algebra " ;  those  wishing  a  more  com- 
plete proof,  in  which  the  general  law  of  coefficients  is  proved 
for  any  two  consecutive  terms,  will  find  it  in  §  447. 

9.  The  proof  of  the  Theorem  of  Undetermined  Coefficients 
given  in  §  396  is  the  same  as  that  given  in  the  "  Essentials  of 
Algebra  " ;  a  more  rigorous  proof  is  given  in  §  450. 

10.  The  author  has  thought  it  best  to  omit  the  proof  of  the 
Binomial  Theorem  for  Fractional  and  Negative  Exponents,  as 
a  rigorous  demonstration  is  beyond  the  capacity  of  pupils  in 
preparatory  schools. 

11.  In  Chapter  XXXIII  will  be  found  Highest  Common 
Factor  and  Lowest  Common  Multiple  by  Division;  and  also 
the  reduction  of  a  fraction  to  its  lowest  terms,  when  the 
numerator  and  denominator  cannot  be  readily  factored  by 
inspection. 

Any  teacher  who  so  desires  can  take  this  work  in  connection 
with  Chapters  IX  and  X. 

Chapter  XXXIII  also  contains  the  proof  of  (1),  §  235,  for 
all  values  of  m  and  n  (§  445);  and  the  reduction  of  a  frac- 
tion whose  denominator  is  irrational  to  an  equivalent  fraction 


PREFACE  V 

having  a  rational  denominator,  when  the  denominator  is  the 
sum  of  a  rational  expression  and  a  surd  of  the  nth  degree,  or 
of  two  surds  of  the  nth  degree  (§  446). 

An  important  feature  of  the  work  is  the  prominence  given 
to  graphical  methods;  in  Chapter  XIII  will  be  found  the 
graph  of  a  linear  equation  with  two  unknown  numbers, 
and  also  of  a  linear  expression  involving  one  unknown 
number. 

In  §§  184,  185,  and  186  will  be  found  the  graphical  repre- 
sentations of  the  solutions  of  simultaneous  linear  equations, 
including  inconsistent  and  indeterminate  equations. 

The  subject  is  taken  up  for  quadratic  equations  in  §§  303  to 
305,  and  314  to  316. 

To  meet  the  demands  of  many  schools,  a  number  of  physi- 
cal problems  have  been  introduced;  these  will  be  found  at 
the  end  of  Exercises  62,  128,  129,  and  145. 

At  the  end  of  the  chapter  on  Variation  will  be  found  a  set 
of  problems  in  physics  in  which  the  principles  of  variation 
are  employed ;  and  also  several  illustrations  of  the  application 
of  graphs  in  physics.  All  the  above  work  in  physics  has  been 
prepared  by  Professor  Eobert  A.  Milliken,  of  the  University 
of  Chicago. 

In  nearly  every  set  of  numerical  equations,  beginning  with 
Exercise  5S,  will  be  found  examples  in  which  other  letters 
than  X,  y,  and  z  are  used  to  represent  unknown  numbers. 

The  examples  and  problems  are  about  4000  in  number; 
and  no  example  is  a  duplicate  of  any  in  the  author's  "Aca- 
demic Algebra"  or  "Essentials  of  Algebra." 

There  is  throughout  the  work  a  much  greater  variety  of 
examples  than  in  the  above  treatises. 

An  important  and  useful  feature  of  the  work  is  the  Index, 


vi  PREFACE 

which,  contains  references  to  all  operations  and  important 
definitions. 

To  meet  the  wants  of  the  most  advanced  schools,  the  author 
has  introduced  nine  additional  chapters/ which  will  cover  the 
entrance  requirements  in  algebra  at  any  American  college  or 
scientific  school. 

These  additional  chapters  are: 

XXXIII.  The  Fundamental  Laws  for  Addition  and  Multipli- 

cation. 

XXXIV.  Additional  Methods  in  Factoring. 
XXXV.     Mathematical  Induction. 

XXXVI.  Equivalent  Equations. 

XXXVII.  Graphical  Eepresentation  of  Imaginary  Numbers. 

XXXVIII.  Indeterminate  Forms. 

XXXIX.  Horner's  Synthetic  Division. 

XL.  Permutations  and  Combinations. 

XLI.  Exponential  and  Logarithmic  Series. 

The  author  desires  to  express  his  thanks  to  the  many 
teachers  in  secondary  schools,  whose  suggestions  in  the  prepa- 
ration of  the  work  have  been  of  the  greatest  service. 

WEBSTER  WELLS. 
Boston,  1906. 


/ 


CONTENTS 

CHAPTER  PAGE 

I.     Definitions  and  Notation 1 

Symbols 1 

Equations .        .  2 

Axioms 2 

Solution  of  Problems  by  Algebraic  Methods  .         .        .3 

Algebraic  Expressions 9 

11.     Positive  anb  Negative  Numbers 11 

Addition  of  Positive  and  Negative  Numbers  ...  12 

Multiplication  of  Positive  and  Negative  Numbers  .        .  14 

III.  Addition  and  Subtraction  of  Algebraic  Expressions  .  17 

Addition  of  Monomials 18 

Addition  of  Polynomials 21 

Subtraction  of  Monomials       .         .         .        ...  24 

Subtraction  of  Polynomials    .        c        ....  26 

Parentheses     . 28 

IV.  Multiplication  of  Algebraic  Expressions       ...  32 

Multiplication  of  Monomials 33 

Multiplication  of  Polynomials  by  Monomials          .        .  84 

Multiplication  of  Polynomials  by  Polynomials       .        .  35 

V.  Division  of  Algebraic  Expressions 42 

Division  of  Monomials    .         .        .         .         .         .         .43 

Division  of  Polynomials  by  Monomials  ....  45 

Division  of  Polynomials  by  Polynomials         .         .         .  •  46 

VI.  Integral  Linear  Equations       .         .         .         .         .         .61 

Principles  used  in  solving  Integral  Equations          .         .  52 

Solution  of  Integral  Linear  Equations    ....  54 
Problems  leading  to  Integral  Linear  Equations  with  One 

Unknown  Number 57 

VII.     Special  Methods  in  Multiplication  and  Division          .  63 


VIII.     Factoring 74 

Miscellaneous  and  Review  Examples     ....  91 

Solution  of  Equations  by  Factoring        ....  94 
vii 


Vlll 


CONTENTS 


CHAPTER 

IX. 


X. 


Highest  Common  Factor.     Lowest  Common  Multiple 
Highest  Common  Factor 
Lowest  Common  Multiple 


j 


\      XI. 


Fractions 

Reduction  of  Fractions 

Addition  and  Subtraction  of  Fractions 

Multiplication  of  Fractions    . 

Division  of  Fractions     . 

Complex  Fractions 

Miscellaneous  and  Review  Examples 

Fractional  and  Literal  Linear  Equations 
Solution  of  Fractional  Linear  Equations 
Solution  of  Literal  Linear  Equations     . 
Solution  of  Equations  involving  Decimals 
Problems  involving  Linear  Equations   . 
Problems  involving  Literal  Equations   . 

Simultaneous  Linear  Equations 

Elimination  by  Addition  or  Subtraction 
Elimination  by  Substitution 
Elimination  by  Comparison  . 
Simultaneous  Linear  Equations  containing 

Two  Unknown  Numbers    . 
Problems  involving  Simultaneous  Linear  Equati 

XIII.  Graphical  Representation 

Graph  of  a  Linear  Equation  involving  Two 

Numbers 

Intersections  of  Graphs 

XIV.  Inequalities 


XIL 


XV.     Involution 


XVL 


Evolution 

Evolution  of  Monomials 

Square  Root  of  a  Polynomial 

Square  Root  of  an  Arithmetical  Numbet 

Cube  Root  of  a  Polynomial    . 

Cube  Root  of  an  Arithmetical  Number 


XVII. 


Theory  of  Exponents 

Miscellaneous  Examples 


More  than 


Unknown 


CONTENTS 


IX 


/ CHAPTER  PAGE 

\j  XVIII.     Surds 222 

Reduction  of  a  Surd  to  its  Simplest  Form  .         .        .  222 

Addition  and  Subtraction  of  Surds     ....  226 
To  reduce  Surds  of  Different  Degrees  to  Equivalent 

Surds  of  the  Same  Degree 227 

Multiplication  of  Surds 228 

Division  of  Surds 230 

Involution  of  Surds 231 

Evolution  of  Surds 233 

Reduction  of  a  Fraction  whose  Denominator  is  Irra- 
tional to  an  Equivalent  Fraction  having  a  Rational 

Denominator 233 

Properties  of  Quadratic  Surds 237 

Imaginary  Numbers 242 

XIX.     Quadratic  Equations 248 

Pure  Quadratic  Equations 248 

Affected  Quadratic  Equations 250 

Problems  involving  Quadratic  Equations  with   One 
/  Unknown  Number  .         .         ..        .         .         .         .261 

XX.     Equations  solved  like  Quadratics       ....  268 

XXI.     Theory  of  Quadratic  Equations 273 

Factoring 276 

Graphical  Representation  of  Quadratic  Expressions 

with  One  Unknown  Number 283 

XXII.     Simultaneous  Quadratic  Equations      ....  286 
Graphical  Representation  of  Simultaneous  Quadratic 

Equations  with  Two  Unknown  Numbers         .         .  300 

XXIII.  Variables  and  Limits 304 

XXIV.  Indeterminate  Equations 308 

XXV.     Ratio  and  Proportion 312 

Properties  pf  Proportion 313 

XXVI.     Variation 321 

XXVII.  Progressions 331 

Arithmetic  Progression       .......  331 

Geometric  Progression 338 

Harmonic  Progression 346 

XXVIII.  The  Binomial  Theorem 350 

Positive  Integral  Exponent 350 


CONTENTS 


CHAPTEK 

XXIX. 


XXX. 
XXXI. 


XXXII. 


XXXIII. 

XXXIV. 

XXXV. 

XXXVI. 

XXXVII. 

XXXVIII. 

XXXIX. 

XL. 

XLI. 


Index 


Undetermined  Coefficients 

Convergency  and  Divergency  of  Series  . 
The  Theorem  of  Undetermined  Coefficients 
Expansion  of  Fractions  .... 
Expansion  of  Surds         .... 


PAGE 

.  357 

.  358 

.  360 

.  361 

.  363 

Partial  Fractions 364 

Reversion  of  Series 370 

The  Binomial  Theorem       .         .         .         .         .         .  372 

Fractional  and  Negative  Exponents        .         .         .  372 

Logarithms  .........  376 

Properties  of  Logarithms        .....  378 

Use  of  the  Table 383 

Applications    ........  388 

Miscellaneous  Examples 391 

Exponential  Equations 393 

Miscellaneous  Topics         ......  395 

Highest   Common   Factor    and   Lowest   Common 

Multiple  by  Division 395 

Proof  of  (1),  §  235,  for  All  Values  of  m  and  u       .  405 
The  Binomial  Theorem  for  Positive  Integral  Expo- 
nents     407 

The  Theorem  of  Undetermined  Coefficients    .         .  408 


The  Fundamental  Laws  for  Addition  and  Multi- 
plication   .... 


Additional  Methods  in  Factoring 
Symmetry       .... 

Mathematical  Induction    . 

Equivalent  Equations 

Graphical  Representation  of  Imaginary 

Indeterminate  Forms 

Horner's  Synthetic  Division    . 

Permutations  and  Combinations 

Exponential  and  Logarithmic  Series 
The  Exponential  Series  . 


The  Logarithmic  Series  . 
Calculation  of  Logarithms 


Numbers 


410 

413 
416 

422 

426 

434 

438 

441 

445 

453 
453 

455 
456 

459 


ALGEBRA  FOR 
SECONDARY  SCHOOLS 


ALGEBRA 


I.    DEFINITIONS  AND  NOTATION 

1.  In  Algebra,  the  operations  of  Arithmetic  are  abridged 
and  generalized  by  means  of  Symbols. 

SYMBOLS  REPRESENTING  NUMBERS 

2.  The  symbols  usually  employed  to  represent  numbers  are 
the  Arabic  Numerals  and  the  letters  of  the  Alphabet. 

The  numerals  represent  known  or  determinate  numbers. 
The  letters  represent  numbers  which  may  have  any  values 
whatever,  or  numbers  whose  values  are  to  be  found. 

SYMBOLS  REPRESENTING  OPERATIONS 

3.  The  Sign  of  Addition,  +,  is  read  "plus J' 

Thus,  a-{-b  signifies  that  the  number  represented  by  b  is  to 
be  added  to  the  number  represented  by  a ;  a  -\-b  -{-  c  signifies 
that  the  number  represented  by  b  is  to  be,  added  to  the  number 
represented  by  a,  and  then  the  number  represented  by  c  added 
to  the  result ;  and  so  on. 

The  result  of  addition  is  called  the  Sum. 

We  shall  use  the  expression  "  the  number  a,"  or  simply  "  a,"  to  signify 
"  the  number  represented  hy  a^^''  etc. 

4.  The  Sign  of  Subtraction,  — ,  is  read  "  minus.^^ 

Thus,  a  —  b  signifies  that  the  number  b  is  to  be  subtracted 
from  the  number  a ;  a—  b  —  c  signifies  that  b  is  to  be  sub- 
tracted from  a,  and  then  c  subtracted  from  the  result;   and 

so  on.  '    *  '.  ^  ^,  =    '  ^"^  !" 


2  ALGEBRA 

5.  The  Sign  of  Multiplication,  X,  is  read  ^' times, ^'  or 
"  multiplied  by." 

Thus,  a  X  b  signifies  that  the  number  a  is  to  be  multiplied 
by  the  number  b;  a  xb  x  c  signifies  that  a  is  to  be  multiplied 
by  b,  and  the  result  multiplied  by  c ;  and  so  on. 

The  sign  of  multiplication  is  usually  omitted  in  Algebra, 
except  between  two  numbers  expressed  in  Arabic  numerals. 

Thus,  2  X  signifies  2  multiplied  by  x ;  but  the  product  of 
2  by  3  could  not  be  expressed  23. 

6.  The  Sign  of  Division,  -;-,  is  read  '^divided  by.^^ 

Thus,  a-7-b  signifies  that  the  number  a  is  to  be  divided  by 
the  number  b. 

The  division  of  a  by  6  is  also  expressed  -• 

EQUATIONS 

7.  The  Sign  of  Equality,  =,  is  read  "  equals. ^^ 

Thus,  a  =  b  signifies  that  the  number  a  equals  the  number  b. 

8.  An  Equation  is  a  statement  that  two  numbers  are  equal. 
The  Jirst  member  of  an  equation  is  the  number  to  the  left 

of  the  sign  of  equality,  and  the  second  member  is  the  number 
to  the  right  of  that  sign. 

Thus,  in  the  equation  2  ic  —  3  =  5,  the  first  member  is  2  ic  —  3, 
and  the  second  member  5. 

AXIOMS 

9.  An  Axiom  is  a  truth  which  is  assumed  as  self-evident. 
Algebraic  operations  are  based  on  the  following  axioms : 

1.  Any  number  equals  itself. 

2.  Any  number  equals  the  sum  of  all  its  parts. 

3.  Any  number  is  greater  than  any  of  its  parts. 

4.  Two  numbers  which  are  equal  to  the  same  number,  or  to 
equal  ^lumbers,  are.  equal. 


DEFINITIONS  AND  NOTATION  3 

5.  If  the  same  number,  or  equal  numbers,  be  added  to  equal 
numbers,  the  resulting  7iumbers  2vill  be  equal. 

6.  If  the  same  number,  or  equal  numbers,  be  subtracted  from 
equal  numbers,  the  resulting  numbers  imll  be  equal. 

7.  If  equal  numbers  be  multiplied  by  the  same  number,  or 
equal  numbers,  the  resulting  numbers  will  be  equal. 

8.  If  equal  numbers  be  divided  by  the  same  number,  or  equal 
numbers,  the  resulting  numbers  will  be  equal. 

SOLUTION  OF  PROBLEMS  BY  ALGEBRAIC  METHODS 

10.  The  following  examples  will  illustrate  the  use  of  Alge- 
braic symbols  in  the  solution  of  problems. 

The  utility  of  the  process  consists  in  the  fact  that  the  un- 
known numbers  are  represented  by  symbols,  and  that  the 
various  operations  are  stated  in  Algebraic  language. 

1.  The  sum  of  two  numbers  is  30,  and  the  greater  exceeds 
the  less  by  4 ;  what  are  the  numbers  ? 

We  will  represent  the  less  number  \)j  x. 
Then  the  greater  will  be  represented  by  x  4-  4. 

By  the  conditions  of  the  problem,  the  sum  of  the  less  number  and  the 
greater  is  30  ;  this  is  stated  in  Algebraic  language  as  follows  : 

x  +  ic  +  4  =  30.  (1) 

Now,  x-\-x  =  xx'2\  for  to  multiply  an  arithmetical  number  by  2,  we 
add  it  twice. 

Again,  x  x  2  =  2  x  aj,  or  2  x  (§  5)  ;  for  the  product  of  two  arithmeti- 
cal numbers  is  the  same  in  whichever  order  they  are  multiplied. 

Therefore,  x  +  x  =  2  x  ;  and  equation  (1)  can  be  written 

2  X  +  4  =  30. 

The  members  of  this  equation,  2  x  +  4  and  30,  are  equal  numbers  ;  if 
from  each  of  them  we  subtract  the  number  4,  the  resulting  numbers  will 
be  equal  (Ax.  6,  §  9). 

Therefore,  2  x  =  30  -  4,  or  2  x  =  26. 

Dividing  the  equal  numbers  2  x  and  26  by  2  (Ax.  8,  §  9),  we  have 

x  =  13. 
Hence,  the  less  number  is  13,  and  the  greater  is  13  +  4,  or  17. 


4  ALGEBRA 

The  written  work  will  stand  as  follows : 

Let  X  =  the  less  number. 

Then,  a;  +  4  =  the  greater  number. 

By  the  conditions,        jc  +  x  +  4  =  30,  or  2  x  +  4  =  30. 

Whence,  2  a;  =  26. 

Dividing  by  2,  x  =  13,  the  less  number. 

Whence,  x  +  4  =  17,  the  greater  number. 

2.  The  sum  of  the  ages  of  A  and  B  is  109  years,. and  A  is 
13  years  younger  than  B;  find  their  ages. 

Let  X  represent  the  number  of  years  in  B's  age. 
Then,  x  —  13  will  represent  the  number  of  years  in  A's  age. 
By  the  conditions  of  the  problem,  the  sum  of  the  ages  of  A  and  B  is 
109  years. 

Whence,  a;  -  13  +  x  =  109,  or  2  cc  -  13  =  109. 

Adding  13  to  both  members  (Ax.  5,  §  9), 

2x  =  122. 

Dividing  by  2,  x  =  61,  the  number  of  years  in  B's  age. 

And,  X  —  13  =  48,  the  number  of  years  in  A's  age. 

The  written  work  will  stand  as  follows : 

Let  X  =  the  number  of  years  in  B's  age. 

Then,  x  —  13  =  the  number  of  years  in  A's  age. 

By  the  conditions,  x  -  13  +  x  =  109,  or  2  x  -  13  =  109. 

Whence,  2  x  =  122. 

Dividing  by  2,  x  =  61,  the  number  of  years  in  B's  age. 

Therefore,  x  —  13  =  48,  the  number  of  years  in  A's  age. 

It  must  be  carefully  borne  in  mind  that  x  can  only  represent  an  abstract 
number. 

Thus,  in  Ex.  2,  we  do. not  say  "let  x  represent  B's  ag^e,"  but  "let  x 
represent  the  number  of  years  in  B's  age." 

3.  A,  B,  and  C  together  have  $  66.  A  has  one-half  as  much 
as  B,  and  C  has  3  times  as  much  as  A.     How  much  has  each  ? 


DEFINITIONS  AND  NOTATION  5 

Let  X  =  the  number  of  dollars  A  has. 

Then,  2  a;  =  the  number  of  dollars  B  has, 

and  3  a;  =  the  number  of  dollars  C  has. 

By  the  conditions,  a;  +  2a;4-3a:  =  66. 

But  the  sum  of  x,  twice  ic,  and  3  times  x  is  6  times  a;,  or  6  x. 

Whence,  Qx  =  m. 

Dividing  by  6,  a;  =  11,  the  number  of  dollars  A  has. 

Whence,  2  x  =  22,  the  number  of  dollars  B  has, 

and  3  a;  =  33,  the  number  of  dollars  C  has. 

(By  letting  x  represent  the  number  of  dollars  A  has,  in  Ex.  3,  we  avoid 
fractions.) 

EXERCISE  I 

1.  The  greater  of  two  numbers  is  8  times  the  less,  and 
exceeds  it  by  49 ;  find  the  numbers. 

''  2.  The  sum  of  the  ages  of  A  and  B  is  119  years,  and  A  is 
17  years  older  than  B ;  find  their  ages. 

3.^  Divide  $  74  between  A  and  B  so  that  A  may  receive  $48 
more  than  B. 

/  4.  Divide  $108  between  A  and  B  so  that  A  may  receive 
5  times  as  much  as  B. 

*  5.  Divide  91  into  two  parts  such  that  the  smaller  shall  be 
one-sixth  of  the  greater. 

"^  6.  A  man  travels  112  miles  by  train  and  steamer ;  he  goes 
by  train  54  miles  farther  than  by  steamer.  How  many  miles 
does  he  travel  in  each  way  ? 

y  7.  The  sum  of  three  numbers  is  69;  the  first  is  14  greater 
than  the  second,  and  28  greater  than  the  third.  Find  the  num- 
bers. 

8.  The  sum  of  the  ages  of  A,  B,  and  C  is  134  years ;  B  is 
13  years  younger  than  A,  and  7  years  younger  than  C.  Find 
their  ages. 


6  ALGEBRA 

9.   A  cow  and  sheep  together  cost  f  91,  and  the  sheep  cost 
one-twelfth  as  much  as  the  cow ;  how  much  did  each  cost  ? 

^  10.  Divide  %  6.75  between  A  and  B  so  that  A  may  receive 
one-fourth  as  much  as  B. 

^  11.  A  man  has  %  2.  After  losing  a  certain  sum,  he  finds 
that  he  has  left  20  cents  more  than  3  times  the  sum  which  he 
lost.     How  much  did  he  lose  ? 

12.  A,  B,  and  C  have  together  %  140 ;  A  has  4  times  as 
much  as  B,  and  C  has  as  much  as  A  and  B  together.  How 
much  has  each  ? 

13.  A,  B,  and  C  have  together  $  100 ;  A  has  %  10  less  than 
C,  and  C  has  $  25  more  than  B.     How  much  has  each  ? 

»^'14.  At  an  election  two  candidates,  A  and  B,  had  together 
653  votes,  and  A  was  beaten  by  395  votes.  How  many  did 
each  receive  ? 

■'  15.  A  field  is  7  times  as  long  as  it  is  wide,  and  the  distance 
around  it  is  240  feet.     Find  its  dimensions. 

16.  My  horse,  carriage,  and  harness  are  worth  together 
$  325.  The  horse  is  worth  6  times  as  much  as  the  harness, 
and  the  carriage  is  worth  $  65  more  than  the  horse.  How 
much  is  each  worth  ? 

17.  The  sum  of  three  numbers  is  87 ;  the  third  number  is 
one-eighth  of  the  first,  and  the  second  number  15  less  than  the 
first.     Find  the  numbers. 

18.  At  a  certain  election,  three  candidates,  A,  B,  and  C, 
received  together  436  votes;  A  had  a  majority  over  B  of  14 
votes,  and  over  C  of  3  votes.     How  many  did  each  receive  ? 

19.  The  sum  of  the  ages  of  A,  B,  and  C  is  110  years ;  B's 
age  exceeds  twice  C's  by  12  years,  and  A  is  9  years  younger 
than  B.     Find  their  ages. 


DEFINITIONS  AND  NOTATION  7 

20.  A  pole  77  feet  long  is  painted  red,  white,  and  black; 
the  red  is  one-fifth  of  the  white,  and  the  black  21  feet  more 
than  the  red.     How  many  feet  are  there  of  each  color  ? 

^  21.  Divide  70  into  three  parts  such  that  the  third  part  shall 
be  one-fifth  of  the  first,  and  one-fourth  of  the  second. 

22.  Divide  $  7.55  between  A,  B,  and  C  so  that  C  may 
receive  one-half  as  much  as  A,  and  B  $  2.95  less  than  A  and 
C  together. 

23.  A,  B,  and  C  have  together  $22.50;  B  has  $1.50  more 
than  A,  and  C  has  $  8  less  than  twice  the  amount  that  A  has. 
How  much  has  each  ? 

y  24.  The  profits  of  a  shopkeeper  in  a  certain  year  were  one- 
third  as  great  as  in  the  preceding  year,  and  $  515  less  than  in 
the  following  year.  If  the  total  profits  for  the  three  years 
were  $  2615,  what  were  the  profits  in  each  year  ? 

25.  The  sum  of  four  numbers  is  96.  The  first  is  4  times  the 
fourth,  and  exceeds  the  third  by  20 ;  and  the  second  exceeds 
the  sum  of  the  first  and  fourth  by  4.     Find  the  numbers. 

*  26.  Divide  $  468  between  A,  B,  C,  and  D,  so  that  A  may 
receive  one-fifth  as  much  as  B,  B  one-fifth  as  much  as  C,  and 
C  one-fifth  as  much  as  D. 

DEFINITIONS 

11.  If  a  number  be  multiplied  by  itself  any  number  of  times, 
the  product  is  called  a  Power  of  the  number. 

An  Exponent  is  a  number  written  at  the  right  of,  and  above 
another  number,  to  indicate  what  power  of  the  latter  is  to  be 
taken;  thus, 

a^,  read  "  a  square,^'  or  "  a  second  power,^^  denotes  a  X  a ; 
a^,  read  " a  cube,^^  or  " a  third power,^^  denotes  a  xaxa\ 
a!^,  read  "  a  fourth/'  "  a  fourth  power/'  or  "  a  exponeyit  4," 
denotes  a  x  a  X  ct  x  a,  etc. 


8  ALGEBRA 

If  no  exponent  is  expressed,  thejirst  power  is  understood. 
Thus,  a  is  the  same  as  a\ 

12.   Symbols  of  Aggregation. 

The  parentheses  (  ),  the  brackets  [  ],  the  braces  \  j,  and  the 

vinculum ,  indicate  that  the  numbers  enclosed  by  them  are 

to  be  taken  collectively ;  thus, 


{a+b)xc,  [a  +  6]  X  c,  [a-\-b\  x  c,  and  a-{-h  x  c 

all  indicate  that  the  result  obtained  by  adding  6  to  a  is  to  be 
multiplied  by  c. 

EXERCISE  2 

What  operations  are  signified  by  the  following  ? 


1. 

2. 

2af/. 
m{x-y). 

5. 

x-(y  +  z). 

8. 

\x-yj 

3. 

ab 

6. 

(m  —  7iy. 

9. 

(2a  +  36)(4c- 

-5d). 

4. 

cd 

x  +  (y-z). 

7. 

a      c 
b     d 

10. 

e-S"- 

Write  the  following  in  symbols : 

11.  The  result  of  subtracting  6  times  n  from  5  times  m. 

12.  Three  times  the  product  of  the  eighth  power  of  m  and 
the  ninth  power  of  n. 

V    13.    The  quotient  of  the  sum  of  a  and  6,  divided  by  the  sum 
of  c  and  d. 
*  14.   The  product  of  ^x-\-y  and  z^. 

15.  The  result  of  subtracting  y  —  z  from  x. 

16.  The  product  oi  a—b  and  c  —  d. 

17.  The  result  of  adding  the  quotient  of  m  by  n,  and  the 
quotient  of  x  by  y. 

18.  The  square  of  m  +  n. 
sj  19.   The  cube  of  a  -  6  +  c 


DEFINITIONS  AND  NOTATION  9 

20.  Tlie  fourth  power  of  the  quotient  of  a  divided  by  x. 

21.  The  product  of  the  quotient  of  1  by  a;  and  the  quotient 
of  1  by  y. 

ALGEBRAIC  EXPRESSIONS 

'  13.  An  Algebraic  Expression,  or  simply  an  Expression,  is  a 
number  expressed  in  algebraic  symbols ;  as, 

2,  a,  or  2aj2_3a6-f  5. 

■^14.  The  Numerical  Value  of  an  expression  is  the  result 
obtained  by  substituting  particular  numerical  values  for  the 
letters  involved  in  it,  and  performing  the  operations  indicated. 

1.  Find  the  numerical  value  of  the  expression 

h 
when  a  =  4,  6  =  3,  c  =  5,  and  d  =  2. 

We  have,    4a  +  — -c?3  =  4  x  4 +  ^^-23  =  16  + 10-8  =18. 
0  o 

If  the  expression  involves  parentheses,  the  operations  indi- 
cated within  the  parentheses  must  be  performed  first. 

2.  Find  the  numerical  value,  when  a  =  9,  &  =  7,  and  c  =  4,  of 

We  have,  a  -  6  =  2,  &  +  c  =  11,  a  +  &  =  16,  and  &  -  c  =  3. 
Then  the  numerical  value  of  the  expression  is 

2xll-M  =  22-l^  =  a 
3  3       3 


EXERCISE  3  ,,  ^P   ' ' 

Find  the  numerical  values  of  the  following  when  a  =  6,  &  =  3, 
c  =  4,  d  =  5,  m  =  3,  and  w  =  2 : 

1.    o?h-cd\  2.    2ahcd,  3.   3a6  +  46c-5cd. 


10  ALGEBRA 

4.  orh\  g    ^_^  I  ^. 

5.  a^  +  6«.  'bed 

'         6.    «+A.  10.  ^  +  \-l-\ 

he     ad  abed 

7  l  +  l-l.  11.  ^_i^.         ;/    ^ 

8  i^  12    ^4-^-^ 

*   28d"*  *   a^     b^     c"' 

Find  the  numerical  values  of  the  following  when  a  =  5,  6  =  2, 
c  =  3,  and  d  =  4 : 


13 


2a4- 


^y.  15.  5a'(a-b)-2b\c  +  d). 


2b-\-' 
14.   (a2-62-d^3.  16.   S(a-by-{-3(c  +  dy 

17.  (a-5)2+(2a-3  6)2-(6  +  c)2. 

18.  (2a-&-cH-d)(2a  +  64-c-d). 


9a  — 46  — 3c  a +  6     a  +  c     a  +  ci 

Find  the   numerical  values  of   the   following  when   a  =  |, 
b  =  ^,  c  =  ^,  and  a;  =  4  : 

oi     g  +  c     a  —  c  22     8a  +  66  — 15  c 

*a  — c     a  +  c  'I6a  +  10  6  +  9c 

23.   a^4-(2a  +  36)ar^-(5a-4c)a;  +  |a&c. 
\a     bj         \a     b      cj        abc 


\ 


POSITIVE   AND  NEGATIVE  NUMBERS  11 


II.  POSITIVE  AND  NEGATIVE  NUMBERS 

15.  There  are  certain  concrete  magnitudes  which  are  capa- 
ble of  existing  in  two  opposite  states. 

Thus,  in  financial  transactions,  we  may  have  assets  or  Ua- 
hilities,  and  gains  or  losses;  we  may  have  motion  along  a 
straight  line  in  a  certain  direction,  or  in  the  opposite  direc- 
tion; etc. 

In  each  of  these  cases,  the  effect  of  combining  with  a  mag- 
nitude of  a  certain  kind  another  of  the  opposite  kind,  is  to 
diminish  the  former,  destroy  it,  or  reverse  its  state. 

Thus,  if  to  a  certain  amount  of  asset  we  add  a  certain 
amount  of  liability,  the  asset  is  diminished,  destroyed,  or 
changed  into  liability. 

16.  The  signs  +  and  — ,  besides  denoting  addition  and  sub- 
traction, are  also  used,  in  Algebra,  to  distinguish  between  the 
opposite  states  of  magnitudes  like  those  of  §  15. 

Thus,  we  may  indicate  assets  by  the  sign  +,  and  liabilities 
by  the  sign  —  ;  for  example,  the  statement  that  a  man's  assets 
are  —  $  100,  means  that  he  has  liabilities  to  the  amount  of  $  100. 

EXERCISE  4 

1.  If  a  man  has  assets  of  $400,  and  liabilities  of  $600,  how 
much  is  he  worth  ? 

'  2.   If  gains  be  taken  as  positive,  and  losses  as  negative,  what 
does  a  gain  of  —  $  100  mean  ? 

3.  In  what  position  is  a  man  who  is  —  3  miles  north  of  a 
certain  place  ? 

4.  In  what  position  is  a  man  who  is  —  50  feet  west  of  a 
certain  point  ? 

5.  How  many  miles  north  of  a  certain  place  is  a  man  who 
goes  5  miles  north,  and  then  9  miles  south  ? 


12  ALGEBRA 

6.  How  many  miles  east  of  a  certain  place  is  a  man  who 
goes  11  miles  west,  and  then  6  miles  east  ? 

17.  Positive   and  Negative  Numbers. 

If  the  positive  and  negative  states  of  any  concrete  magni- 
tude be  expressed  witJiout  reference  to  the  unit,  the  results  are 
called  positive  and  7iegative  numbers,  respectively. 

Thus,  in  -{-  $5  and  —  $3,  +5  is  a  positive  number,  and  —3 
is  a  negative  number. 

For  this  reason  the  sign  4-  is  called  the  positive  sign,  and 
the  sign  —  the  negative  sign. 

'  If  no  sign  is  expressed,  the  number  is  understood  to  be  posi 
tive ;  thus,  5  is  the  same  as  -f  5. 

■  The  negative  sign  must  never  be  omitted  before  a  negative 
number. 

18.  The  Absolute  Value  of  a  number  is  the  number  taken 

independently  of  the  sign  affecting  it. 
Thus,  the  absolute  value  of  —  3  is  3. 

ADDITION  OF  POSITIVE  AND  NEGATIVE  NUMBERS 

19.  We  shall  give  to  addition  in  Algebra  its  arithmetical 
meaning,  so  long  as  the  numbers  to  be  added  are  positive  integers 
or  positive  fractions. 

We  may  then  attach  any  meaning  we  please  to  addition 
involving  other  forms  of  numbers,  provided  the  new  meanings 
are  not  inconsistent  with  principles  previously  established. 

20.  In  adding  a  positive  number  and  a  negative,  or  two 
negative  numbers,  our  methods  must  be  in  accordance  with 
the  principles  of  §  15. 

If  a  man  has  assets  of  $  5,  and  then  incurs  liabilities  of  $  3, 
he  will  be  worth  $2. 

If  he  has  assets  of  ^  3,  and  then  incurs  liabilities  of  $  5,  he 
will  be  in  debt  to  the  amount  of  $2. 


POSITIVE   AND  NEGATIVE  NUMBERS  13 

If  he  has  liabilities  of  $5,  and  then  incurs  liabilities  of  f  3, 
he  will  be  in  debt  to  the  amount  of  $  8. 

Now  with  the  notation  of  §  15,  incurring  liabilities  of  $3 
may  be  regarded  as  adding  —  $  3  to  his  property. 

Whence,     the  sum  of  +  f  5  and  —  $3  is  +  ^2; 

the  sum  of  —$5  and  +$3  is  —  $2; 

and  the  sum  of  —$5  and  —  $ 3  is  —  $8. 

Or,  omitting  reference  to  the  unit, 

(  +  5)  +  (-3)=:+2; 

(_5)  +  (  +  3)=-2; 

(_5)  +  (-3)=-8. 

To  indicate  the  addition  of  +  5  and  —  3,  they  must  be  enclosed  in 
parentheses  (§  12). 

We  then  have  the  following  rules : 

^  To  add  a  j^ositive  and  a  negative  number,  subtract  the  less 
absolute  value  (§  18)  from  the  greater,  and  prefix  to  the  result 
the  sign  of  the  number  having  the  greater  absolute  value. 

y^  To  add  two  negative  numbers,  add  their  absolute  values,  and 
prefix  a  negative  sign  to  the  result. 

21.   Examples. 

1.  Find  the  sum  of  + 10  and  —  3. 
Subtracting  3  from  10,  the  result  is  7. 
Whence,  (  +  10)  +  (  -  3)  =  +  7. 

2.  Find  the  sum  of  — 12  and  +  6. 
Subtracting  6  from  12,  the  result  is  6. 
Whence,  (_  12)  +  (  +  6)  =-6. 

3.  Add  _  9  and  -  5. 

The  sum  of  9  and  5  is  14. 

Whence,  (_  9)  +  (_  5)  =  _„  14 


14  ALGEBRA 

EXERCISE    5 

Find  the  values  of  the  following : 
</l.   (-6) +  (-7). 
2.    (+8) +  (-3). 


'  (-IH-1 


3.    (-9)+(+5).  9.   f+D  +  f-l 

»/4.  (+4) +  (-11).  V  »y    V   ^, 

5.  (_13)  +  (-18).  10-    (-15i)  +  (+12x). 

6.  (_42)  +  (-f57).  11.    (+17|)  +  (-10A). 
1^7.    (-34) +  (+82).  12.    (-14|)  +  (-21A). 

MULTIPLICATION  OF  POSITIVE  AND  NEGATIVE 
NUMBERS 

22.  If  two  expressions  are  multiplied  together,  the  first  is 
called  the  Multiplicand,  and  the  second  the  Multiplier. 

The  result  of  multiplication  is  called  the  Product. 

23.  We  shall  retain  for  multiplication,  in  Algebra,  its  arith- 
metical meaning,  so  long  as  the  multiplier  is  a  jjositive  integer  or 
a  positive  fraction. 

That  is,  to  multiply  a  number  by  a  positive  integer  is  to  add 
the  multiplicand  as  many  times  as  there  are  units  in  the  mul- 
tiplier. 

For  example,  to  multiply  —  4  by  3,  we  add  —  4  three  times. 

Thus,  (-4)x(+3)  =  (-4)  +  (-4)  +  (-4)  =  -12. 

24.  In  Arithmetic,  the  product  of  two  numbers  is  the  same 
in  whichever  order  they  are  multiplied. 

Thus,  3x4  and  4  x  3  are  each  equal  to  12. 
If  we  could  assume  this  law  to  hold  for  the  product  of  a 
positive  number  by  a  negative,  we  should  have 

(+3)  X  (-4)  =  (-4)  X  (+3)  =  -12  (§  23)  =-  (3  x4). 


POSITIVE   AND  NEGATIVE   NUMBERS  15 

Then,  if  the  above  law  is  to  hold,  we  must  give  the  follow- 
ing meaning  to  multiplication  by  a  negative  number : 

A  To  multiply  a  number  by  a  negative  number  is  to  multiply  it  by 
the  absolute  value  (§  18)  of  the  multiplier,  and  change  the  sign  of 
the  result. 

Thus,  to  multiply  +  4  by  —  3,  we  multiply  +4  by  +3, 
giving  + 12,  and  change  the  sign  of  the  result. 

That  is,  (+  4)  X  (-  3)  =  - 12. 

Again,  to  multiply  —  4  by  —  3,  we  multiply  —  4  by  +3, 
giving  — 12  (§  23),  and  change  the  sign  of  the  result. 

That  is,  (-4)x(-3)  =  +12. 

25.  From  §§23  and  24  we  derive  the  following  rule : 

'^To  multiply  one  iiumber  by  another,  multiply  together  their 
absolute  values. 

X  Make  the  product  plus  when  the  multiplicand  and  multiplier  are 
of  like  sign,  and  minus  whe7i  they  are  of  unlike  sign. 

26.  Examples. 

1.  Multiply  +8  by  -5. 

By  the  rule,         (+  8)  x  (-  5)  =  -  (8  x  5)  =  - 40. 

2.  Multiply  -  7  by  -  9. 

By  the  rule,  (-  7)  x  ( -  9)  =  +  (7  x  9)  =  +  63. 

3.  Find  the  numerical  value  when  a  =  4  and  b  =  —  7,  of 

(a  +  by. 
We  have,  (a  +  6)3  =  (4  -  7)(4  -  7)(4 -7) 

=  (-3)(-3)(-3)  =  -27. 

EXERCISE    6 
Find  the  values  of  the  following : 

1.    (+,5)x(-4).  2.    (-11)  X  (+3). 


16 


3 

ALGEBRA 

v'3.    (_8)x(-7). 
v^4.    (+9)x(-6). 

10. 

(-§)K-ii> 

5.  (-12)  X  (+9). 

6.  (-24)  X  (-5). 

^^ll. 

(-i)K-s> 

7.    (-14)x(+15). 

12. 

(-7|)x(-«). 

8.    (+27)x(-19). 

13. 

(-6f)x(+6f). 

,9.   (-l)x(-l\ 

14. 

(+1H)X(-1H)- 

Find  the  numerical  value  when  a  =  2,  b=  —  4:^  c  =  5,  and 
d=-3  of 

15.   6«.  16.   d"^.  17.   (6(^)2.  18.    (abcf. 

•19.    (2  a- 5  6) (4c +  3  d).  20.    (a-b)(b-c)(c-d). 

v^21.    (a  +  6)(c  +  d)-(a-c)(6-d). 

22.   a26-3  62c-2cU  23.    (a-cY+{b±d}\ 

/24.  3a-6^-5  6V  +  4c^(^^. 


ADDITION  AND  SUBTRACTION  17 


III.    ADDITION  AND  SUBTRACTION  OF 
ALGEBRAIC    EXPRESSIONS.      PARENTHESES 

*i27.  A  Monomial,  or  Term,  is  an  expression  (§  13)  whose 
parts  are  not  separated  by  the  signs  +  or  —  ;  as  2  a?^,  —  3  ab, 
or  5. 

2x^,  —  3  ah,  and  +  5  are  called  the  terms  of  the .  expression 
2a;2_3a6  +  5. 
«^A  Positive  Term  is  one  preceded  by  a  4-  sign ;  as  -\-5a. 

If  no  sign  is  expressed,  the  term  is  understood  to  be  posi- 
tive. 

•''A  Negative  Term  is  one  preceded  by  a  —  sign ;  as  —  3  ah. 
The  —  sign  must  never  be  omitted  before  a  negative  term. 

/  28.  If  two  or  more  numbers  are  multiplied  together,  each  of 
them,  or  the  product  of  any  number  of  them,  is  called  a  Factor 
of  the  product. 

Thus,  a,  h,  c,  ah,  ac,  and  he  are  factors  of  the  product  ahc. 

*"  29.  Any  factor  of  a  product  is  called  the  Coefficient  of  the 
product  of  the  remaining  factors. 

Thus,  in  2  ah,  2  is  the  coefficient  of  ah,  2  a  of  h,  a  oi  2  h,  etc. 

V  30.  If  one  factor  of  a  product  is  expressed  in  Arabic  numer- 
als, and  the  other  in  letters,  the  former  is  called  the  numerical 
coefficient  of  the  latter. 

Thus,  in  2  ah,  2  is  the  numerical  coefficient  of  ah. 

If  no  numerical  coefficient  is  expressed,  the  coefficient  1  is 
understood ;  thus,  a  is  the  same  as  1  a. 

31.   By  §  25,  (-3)  X  a  =  -  (3  X  a)  =  -3a. 

That  is,  —  3  a  is  the  product  of  —  3  and  a. 

Then,  —  3  is  the  numerical  coefficient  of  a  in  —  3  a. 
/Thus,  in  a  negative  term  as  in  a  positive,  the  numerical  coeffir 
dent  includes  the  sign. 


18  ALGEBRA 

^  32.  Similar  or  Like  Terms  are  those  which  either  do  not 
differ  at  all,  or  differ  only  in  their  numerical  coefficients ;  as 
2  x^y  and  —  7  x^y. 

•"  Dissimilar  or  Unlike  Terms  are  those  which  are  not  similar ; 
as  3  x^y  and  3  xy"^. 

ADDITION  OF  MONOMIALS 

33.   The  sum  of  a  and  h  is  expressed  a  +  6  (§  3). 

V  34.  We  define  Subtraction,  in  Algebra,  as  the  process  of 
finding  one  of  two  numbers,  when  their  sum  and  the  other 
number  are  given. 

The  Minuend  is  the  sum  of  the  numbers. 

The  Subtrahend  is  the  given  number. 

The  Remainder  is  the  required  number. 

35.  The  remainder  when  h  is  subtracted  from  a  is  expressed 
a  -  6  (§  4). 

Since  the  sum  of  the  remainder  and  the  subtrahend  gives 
the  minuend  (§  34),  we  have 

a—h-\-h  =  a. 

^  Hence,  if  the  same  number  he  both  added  to,  and  subtracted 
from,  another,  the  value  of  the  latter  is  not  changed. 

^  36.  It  follows  from  §  35  that  terms  of  equal  absolute  value, 
but  opposite  sign,  in  an  expression,  may  be  cancelled. 

37.  We  will  now  show  how  to  find  the  sum  of  a  and  —  b. 
By  §35,  a  +  (-6)  =  a  +  (-5)  +  6-5;  (1) 

for  adding  and  subtracting  b  does  not  alter  a  +  (—  6). 
But  by  §20,  {-b)  +  b  =  0', 

for  —  b  and  b  are  numbers  of  the  same  absolute  value,  but 
opposite  sign. 

Therefore,  a+(—  6)  =  a-  —  6; 

for  the  other  terms  in  the  second  member  of  (1)  cancel  each 
other. 


ADDITION   AND   SUBTRACTION  19 

*^38.  It  follows  from  §§33  and  37  that  the  addition  of  mono- 
mials is  effected  by  uniting  their  terms  with  their  respective  signs 
Thus,  the  sum  of  a,  —b,c,  —  d,  and  —  e  is 
a—b-\-c  —  d  —  e. 

-^39.   We  assume  that  the  terms  can  be  united  in  any  order j 
provided  each  has  its  proper  sign. 

Hence,  the  result  of  §  38  can  also  be  expressed 

c-{-a  —  e  —  d—bf  — d  —  &  +  c  —  e-f-a,  etc. 
This  law  is  called  the  Commutative  Law  for  Addition  ;  compare  §  451. 

^  40.  To  multiply  5  +  3  by  4,  we  multiply  5  by  4,  and  then 
3  by  4,  and  add  the  second  result  to  the  first. 

Thus,  (5  +  3)4=5x4  +  3x4. 

We  then  assume  that  to  multiply  a  -\-b  by  c,  we  multiply 
a  by  c,  and  then  b  by  c,  and  add  the  second  result  to  the  first. 

Thus,  (a  -\-  b)c  =  ac  -\-  be. 

This  law  is  called  the  Distributive  Law  for  Multiplication ;  its  proof 
for  the  various  forms  of  number  will  be  found  in  §  455. 

41.   Addition  of  Similar  Terms  (§  32). 

1.  Required  the  sum  of  5  a  and  3  a. 

We  have,  5  a  +  3  a  =  (5  +  3)a  (§40) 

=  8  a. 

2.  Required  the  sum  of  —  5  a  and  —  3  a. 

Wehave,        (_  5a)  +  (- 3a)  =  (- 5)  x  a+(-3)xa  (§31) 

=  [(-5)  +  (-3)]x«  (§40) 

=  (-8)xa  (§20) 

=  -8a.  (§31) 

3.  Required  the  sum  of  5  a  and  —  3  a. 

Wehave,  5  a +(- 3)a  =[5 +(-3)]  x «  (§40) 

=  2  a.  (§  20) 


20  ALGEBRA 

4.  Required  the  sum  of  —5a  and  3a. 

We  have,  (_  5)a  +  3  a  =[(- 5)  + 3]  x  a  (§40) 

=  (-2)x  a  (§20)  =  - 2  a. 

/  Therefore,  to  add  two  similar  terms,  find  the  sum  of  their 
numerical  coefficients  (§§  20,  30,  31),  and  affix  to  the  result  the 
common  letters. 

5.  Find  the  sum  of  2  a,   —  a,  3  a,  —  12  a,  and  6  a. 

Since  the  additions  may  be  performed  in  any  order,  we  may  add  the 
positive  terms  first,  and  then  the  negative  terms,  and  finally  combine 
these  two  results. 

The  sum  of  2  a,  3  a,  and  6  a  is  11  a. 

The  sum  of  —  a  and  —  12  a  is  —  13  a. 

Hence,  the  required  sum  is  11  a  +  (—  13  a),  or  —2  a. 

6.  Add  3{a-h),  -2(a-b),  6(a-b),  and  -4(a-6). 

The  sum  of  3(a  -  b)  and  6(a  -  6)  is  9(a  -  6). 

The  sum  of  —  2(a  —  &)  and  —  4(a  —  &)  is  —  6(a  —  6). 

Then,  the  result  is  [9  +(-^  6)](a  -  6),  or  3(a  -  b). 

If  the  terms  are  not  all  similar,  we  may  combine  the  similar 
terms,  and  unite  the  others  with  their  respective  signs  (§  38). 

7.  Required  the  sum  of  12  a,  —5x,  —  3  /,  —5  a,  Sx,  and 
—  3  a;. 

The  sum  of  12  a  and  —  5  a  is  7  a. 

The  sum  of  -5x,  Sx,  and  -  3  a;  is  0  (§  36). 

Then,  the  required  sum  is  1  a  —  Sy^. 

EXERCISE  7 

Add  the  following : 

^  1.   11  a  and  —6  a.  6.   —  abc  and  12  abc. 

2.  7x  and  -10  a;.  7.   8a^2/'  and  -29  0^^/. 

3.  -4n  and  -9n.  w8.   9(a  +  6)  and  -2(a  +  6). 

4.  —13ab  and  5db.  9.   —IWmn^  and  60a%nl 

(, 

V  5.   -17 a^  and  -loaj^.        ^  10.   8a,  la,  and  -^9a. 


ADDITION   AND   SUBTRACTION  21 

11.  15m,  —m,  —5  771,  and  —12  m. 

12.  16  xyz,  —  4  xyz,  xyz,  and   —  6  xyz. 

13.  Q>{x-y),   -5(x-y),  and  -10(x-y). 

14.  ISn^,   -13  7i2,  2n2,  -  7^^,  and  -147il 

15.  19a^6,  2a^b,   -^a%  -17a%  and  lOa^d. 

16.  7  ax,  —  9  62/j  —  3  aa;,  and  2  6?/. 

rl7.  8  ic,  2;,  —  5  2/,   —llz,  —2x,  and  10  ?/. 

18.  '8  (m  +  7?.),  4  (m  —  n),  —  3  (m  4-  w),  and  —  7  (m  —  ti). 

V  19.  14  a,   —  4  d,  —8  c,  b,  —2  a,  —3  c,  — 15  d,  and  —  c. 

•20.  6  a?,  —7y,5z,Sy,  —4:Z,  —Sx,  —y,  —9z,  and  2x. 

•/"addition  of  polynomials 

42.  A  Polynomial  is  an  algebraic  expression  consisting  of 
more  than  one  term  ;  as  a  +  5,  or  2  a:^  —  ic?/  —  3  2/^. 

A  polynomial  is  also  called  a  multinomial. 

A  Binomial  is  a  polynomial  of  two  terms  ;  as  a  +  6. 

A  Trinomial  is  a  polynomial  of  three  terms  ;  as  a  -h  6  —  c. 

43.  A  polynomial  is  said  to  be  arranged  according  to  the 
descending  powers  of  any  letter,  when  the  term  containing  the 
highest  power  of  that  letter  is  placed  first,  that  having  the  next 
lower  immediately  after,  and  so  on. 

Thus,  ic*  +  3  ar^?/-  2a^/-f  3aj/-4  2/^ 

is  arranged  according  to  the  descending  powers  of  x. 

The  term  —  4  y^^  which  does  not  involve  x  at  all,  is  regarded  as  con- 
taining the  lowest  power  of  x  in  the  above  expression. 

A  polynomial  is  said  to  be  arranged  according  to  the  ascend- 
ing powers  of  any  letter,  when  the  term  containing  the  lowest 
j)Ower  of  that  letter  is  placed  first,  that  having  the  next  higher 
immediately  after,  and  so  on. 

Thus,  x^  +  Z  o^y -2  Q^y^ -\-S  xf  -  A:  y^ 

is  arranged  according  to  the  ascending  powers  of  y. 


22  ALGEBRA 

44.  Addition  of  Polynomials. 

Let  it  be  required  to  add  &  +  c  to  a. 

Since  &  +  c  is  the  sum  of  h  and  c  (§  3),  we  may  add  6  -|-  c  to 
a  by  adding  h  and  c  separately  to  a. 

Then,  a  +  (6  +  c)  =  a  -f-  &  4-  c. 

(To  indicate  the  addition  of  6  +  c,  we  enclose  it  in  parentheses.) 

The  above  assumes  that,  to  add  the  sum  of  a  set  of  terms,  we  add  the 

terms  separately. 

This  is  called  the  Associative  Law  for  Addition;  its  proof  will  be  found 

in  §  452. 

45.  Let  it  be  required  to  add  6  —  c  to  a. 
By  §  37/  5  —  c  is  the  sum  of  h  and  —  c. 

Then,  to  add  h  —  c  to  a,  we  add  h  and  —  c  separately  to  a 
(§  44). 
Whence,  a  +  (6  —  c)  =  a  +  6  —  c. 

46.  From  §§44  and  45  we  have  the  following  rule : 

*'  To  add  a  polynomial,  add  its  terms  with  their  signs  unchanged. 
1.   Add  Qa-Tx",  3a^-2a  +  3/,  smd -2  x^  -  a  -  mn. 

We  set  the  expressions  down  one  underneath  the  other,  similar  terms 
being  in  the  same  vertical  column. 

We  then  find  the  sum  of  the  terms  in  each  column,  and  write  the 
results  with  their  respective  signs  ;  thus, 

6  a  -  7  a;2 

-  2  a  +  3  a;2  +  3  ?/3 

—  a  +  2aj2  —  mn 


a  —  2x'^  +  Sy^  —  mn 


2.   Add    4:x-3a^-ll  +  5x%     12a^- 7-8  a^-15a;,     and 
U-\-6a^-\-10x-9x'. 

It  is  convenient  to  arrange  each  expression  in  descending  powers  of  x 

(§  43) ;  thus, 

5a;3_    3a;2+    4a; -11 

-  8  cc3  +  12  5c2  -  15  X  -    7 

6x^-    9a;2  +  i0a;+14 

Sx^  -      X-    A 


ADDITION   AND   SUBTRACTION  23 

3.  Add      9(a  +  &)  -  8(6  +  c),       _  3(6  +  c)  -  7(c  +  a),      and 
4(c  +  a)-5(a  +  6). 

9(a  +  &)-   8(&  +  c) 

-   3(&  +  c)-7(c  + a) 
-5(«  +  6)  +4(c  +  a) 

4(a  +  6)  -  11(6  +  c)  -  3(c  +  a) 

4.  Add  fa  +  |6  — ic  and  ia-|6  +  fc. 


Ha-Hb  +  ^\c 

EXERCISE  8 

Add  the  following : 

1. 

2. 

«^. 

Sa-7b 

-   6(^-Wy' 

—  17  am  4-   4  6n 

—  5  «  -f  4  5 

9a^+   3/ 

6  am  — 11  bn 

a-26 

- 12  a^  4- 10  2/' 

9  am  + 19  bn 

4.  7  X  +  62/  —  9 2  and  4:X  —  Sy-{-5z. 

5.  4  m^  —  4  mii  +  n^,  m^  +  4  m?i  +  4  n^,  and  —  5  m^  +  5  n^. 

6.  5  a  — 7  b,  4  6-9  c,  and  6  c  — 2  a. 

1^7.  3aj2-2a;2/  +  7?/2,  -5  i^-\-9  xy-10 y%  and  S  x'-S  xy-4:  y\ 

8.  a-9-8a2  +  16a^   5 +  15  a=^- 12  a-2  a^, 

and   Ga^-lOa^  +  lla-ia. 

9.  5(^a  +  b)-4.x(x--y),    -6{a-\-b)-\-Sx(x-y)y 

and   8  (a  +  6)  —  7  a?  (a;  —  y). 

10.   fa  — -1-6  — y\c  and  -fa  +  i6  — -f-c. 

Vll.   5m4-9w  +  4(»,     _3a;-72/-6n,      - 10 2/  +  8a;  +  2m, 
and  n-\-lly  —  7m. 

12.    3iV«^  +  t2/  +  H2  and  3-\a;-|^-i2;. 


24  ALGEBRA 

13.  U(x-{-y)-17(y  +  z),   4(y  +  z)-9(z-{-x\ 

and    —3(x-\-y)  —  7{z-\-x). 

14.  6c  +  2a-3b,   4.d-7c-^12a,   Sb-5d  +  c, 

and    -10  a -11  6  +  9  d 

n5.     -7(a-5)2  +  8(a-6)+2,  4(a -  6)2 - 5 (a - 6), 
and   3(a- 6)2-9. 

16.  8a3-lla-7a2,  2a-6a2  +  10,  -5  +  4a3  +  9a, 

and   13a'-5-12^ 

17.  a;22,  +  2a;/  +  3ic^    3  a;/ +  4  ^/^  _  5  ^^^^    6a^ -\-5f -7  xy', 

and    _82/'  +  9i^2/-7a^. 
'^-IS.   lla^-13  +  4a;3  +  5a;,   -14a;  +  2a^4-7  +  12a^, 

Sx'-3x-10-{-6x%   and    1- 15cc2  +  9a;- 16  a^l 
19.   |a2-|a-|,   -ia^  +  a  +  |,  and  --Ja2_  e  ^_^|., 

•^20.    5mhi-n^-4.m^-\-2m7i%    7  mn^ -lSm^n  +  2m^-97i^, 
-  15  mw^  +  3  m^n  +  16  71^  +  8  m^ 
and  —  5  m^  +  3  mii^  —  6  ii^  +  10  m^w. 

i^l.    -6n3 +  271-12-15  7^2,   -14  +  7n-n2-97i3, 

6?i2_|-l3n3^3_ll^^  and  8-167i  +  107i2  +  4wl 


i 


SUBTRACTION  OF  MONOMIALS 

47.  The  remainder  when  6  is  subtracted  from  a  is  expressed 
a-6(§4). 

We  will  now  show  how  to  subtract  —  6  from  a. 

By  §  34,  the  sum  of  the  remainder  and  the  subtrahend  equals 
the  minuend. 

Then,  the  required  remainder  must  be  an  expression  such 
that,  when  it  is  added  to  —  6V^he  sum  shall  equal  a. 

But  if  a  +  6  is  added  to  —  6,  the  sum  is  a  (§  35). 

Therefore,  the  required  remainder  is  a  +  6. 

That  is,  a-(-6)  =  a  +  6. 

48.  From  §  47,  we  have  the  following  rule : 

To  subtract  a  monomial,  change  its  sign,  and  add  the  result  to 
the  minuend. 


ADDITION   AND   SUBTRACTION  25 

1.  Subtract  5  a  from  2  a. 

Changing  the  sign  of  the  subtrahend,  and  adding  the  result  to  the  min- 
uend, 

2  a  -  5  rt  =  2  a  +  (- 5  a)  = -3  a  (§  41). 

2.  Subtract  —  2  a  from  5  a, 

5«^(-2a)  =  5a  +  2a  =  7a. 

3.  Subtract  —b  a  from  —2  a. 

-2a-(-5a)=-2a-f5«  =  3a. 

4.  Subtract  5{x-\-y)  from  —2(x-\-y). 

-  2(a-- +  ?/)- 5(x +  ?/)  =  - 7(x  +  ?/). 

The  pupil  should  endeavor  to  put  down  the  results,  in  examples  like  the 
above,  without  writing  the  intermediate  step ;  changing  the  sign  of  the 
subtrahend  mentally,  and  adding  the  result  to  the  minuend. 

5.  From  —  23  a  take  the  sum  of  19  a  and  —5  a. 

It  is  convenient  to  change  the  sign  of  each  expression  which  is  to  he 
subtracted,  and  then  add  the  results. 

We  then  have  —  23  a  —  19  a  +  5  a,  or  —  37  a. 

EXERCISE  9 
Subtract  the  following : 
1.   9  from  3.  4.   -5  from  12.  1,   -3  from  3^. 

V2.   2  from  -  6.  5.   42  from  15.  8.   -  |f  from  -  f . 

3.   - 16  from  - 10.   v'6.   _  28  from  -  61.     9.  lOf  from  -  3f 
10.  11.  vl2.  13.  ^li. 

14  a  4  a;  —^a^  — 15  mn  —   7  x'y 

8a  -11a;  4a^  -   1  mn  — 12  x'y 

15.  5  he  from  he.  19.  19  {a-h)  from  17  {a-h). 

16.  a;2/2  from  -  8  a;?/2.  20.  -  18  a%(^  from  -  45  a^ftc^. 

17.  25  aV  from  13  a'x^  '  21.  From  7  x  take  -  11  .y. 

18.  -  40  ahc  from  -  23  ahc.  22.  From  -2a^  take  5  w^. 


26  ALGEBRA 

23.    From  the  sum   of   18  ab  and  —9ab  take  the  sum  of 

—  21  ab  and  11  ab. 

^'24.    From  the  sum  of  —13n^  and  24:  n^  take  the  sum  of 
46  n^  and  —  19  n^. 

V^  25.    From  the  sum  of  16  xy^  and  —  37  xy^  take  the  sum  of 

—  29  iC2/^,  34  a;?/^,  and  —  47  iC2/^. 

J\       SUBTRACTION  OF  POLYNOMIALS 

49,  Since  a  polynomial  may  be  regarded  as  the  sum  of  its 
separate  terms  (§  38),  we  have  the  following  rule : 

To  subtract  a  polynomial,  change  the  sign  of  each  of  its  terms, 
and  add  the  result  to  the  minuend. 

1.  Subtract  7  a^^  _  9  a-b  +  8  6^  from  5  a^  -  2  a^^  +  4  ab\ 

It  is  convenient  to  place  the  subtrahend  under  tlie  minuend,  so  that 
similar  terms  shall  be  in  the  same  vertical  column. 

We  then  mentally  change  the  sign  of  each  term  of  the  subtrahend,  and 
add  the  result  to  the  minuend  ;  thus, 

,  5  a3  _  2  a26  +  4  a62 

-  9  a%  +  7  a62  ^_  8  53 
5a3  +  1  a^b  -  3  a&2  _  g  fts 

2.  Subtract  the  sum  of  9x^  —  %x-\-x^  and  ^  —  a?-\-x  from 
Q,^-lx-4. 

We  change  the  sign  of  each  expression  which  is  to  be  subtracted,  and 

add  the  results.  ^    „  „         m 

Qx^  —  7  a;  —  4 

-  x^  -  9  a;2  +  8  x 

+    a;2  -    a;  -  5 
6x3-8x2  Tq 


EXERCISE  iO 

Subtract  the  following: 

1.  V  2.  3. 

a^  4- 13  oj  —  11  —  2  m^  —  4  mn  4-   9  n^  a6  -}-  6c  +  ca 

—  3ar'-|-    Q>x  —    5  8m^  —  7 m?i  +.  14 n^  ah  —  bc-{- ca 


ADDITION   AND   SUBTRACTION  27 

4.  From  Sx  +  2y  —  7z  subtract  8x  —  2y  +  7z. 

5.  From  Aa^- 5a^ -15a-6  take  a^ -12a^ -3a  +  ll. 

6.  From  7a  — 9c  — 6  subtract  — 5c  +  12a  — 8  6. 

7.  Subtract  —5(x+y)-\-9(x—y)  from  7  (x+y)—6{x—y). 

8.  Take  49  x^  + 16  m^  -  56  mx  from  25  m^  +  36  a^  -  60  mx. 

9.  By  how  much  does  15  ar^  +  6  a^y  —  4  a??/^  — 11 2/^ 

exceed  Sa^-9x^y -i-Uxy'- -Sf? 

•lO.   Take  8a^-12a26  4-6a62-63 

from  a''  -  6  a'b  +  12  a^^  _  8  b^ 

V 11.   What  expression  must  be  added  to  3  a^— a;  +  5  to  give  0  ? 

12.  By  how  much  does  2  m  —  4  m^  — 15  + 17  m^ 

exceed  —  9  +  6  m^  —  11  m  — 14  m^  ? 

13.  From  a;  +  15  ar^- 18  subtract  -  2  ar^- 13  +  41  a^. 

14.  Take  3&-16d  +  7a-10c  from  -13c-hl4 a-5d^9  6. 

15.  Subtract  12x  —  7  7i  —  6y  from  11  n  +  3  m  —  8  x. 

^16.   From  7  7r-5+20n3+13n  take  -9-14  n^ +16n+5  Til 

17.  From  |.  ^^  _  _i_  5  .f  1 9  c  subtract  ia  +  |&  — |c. 

18.  Subtract  15a-21a2  +  17  from  -  12a^  +  22a3-9a 

19.  Take  a^-ea^- i5a2_8a  +  4 

from  7a^  +  3a''-5a2-lla-9. 

20.  From  im  — Jn  +  ^p  take  fm  — J?i  +  -Jp. 

^1.    From  71^  -  10  A' -  nV  +  8  nar5  + 3  a;^ 

take  5n^  +  4n3a;-9n2aj2  +  2wa.'3-12a;*. 

22.  Take  18a;^-8a.-  +  6a^4-12-8a.'3 

from  -10a;3^2-15a;2  +  llar'-4a;. 

23.  Take  a-^  - 1 0  a%^  + 13  a'b^  -  7  a?/  -  5  6^ 
from  9  a^  +  3  a^6  H-  6  cfb'  -  arb^  - 16  6«. 

4.   From  the  sum  of  2x^—bxy- y'^  and  7 y? —  3xy-\-9y^ 
subtract  4  a^  —  6  a;?/  +  8  ?/^.  .V  \ 

V  25.    From  0  subtract  the  sum  of  4  (jl-  and  3  a  —  5  a^  —  1. 


V 


28  ALGEBRA 

26.  From  7 x  —  5z  —  3y  subtract  the  sum  of  S y  -\-  2 x  —11  z 

and  6 z —  12 y-^4:X. 

27.  From  671^  —  671  — 11  subtract  the  sum  of  2  n^  —  4 n  —  3, 

7w2-10?i  +  4,  and  -3n2  +  8n-12. 
•  28.   From  the  sum  of  36  +  2 a  —  4c  and  9G-\-3b  —  5d 
subtract  the  sum  of  —6  d  — 7  a 
and  8a  — 7d  +  96  +  5c. 
29.   From  the  sum  of  4.a-l-\-5a^-Sa%  11-9 a" +3 a^ -7 a, 
and  3a^-7+10a-a^ 

\  subtract  -4.a^-{-9 a-Qa' +  2. 

30.   From  the  sum  of  7 a^-4:X^-\-6x  and  3x^-10x-5 
take  the  sum  of— 5a^-f4a;4-12 
and  8a;3_ii^2_2. 

PARENTHESES 
50.   Removal  of  Parentheses. 

By  §  45,  a  +  (6  —  c)  =  a  +  &  —  c. 

Hence,  jparentlieses  preceded  by  a  -\-  sign  may  he  removed 
without  changing  the  signs  of  the  te7'ms  enclosed. 

Again,  by  §  49,     a  —  {h  —  c)  =  a  —  h-{-G. 

Hence,  pa7-entheses  preceded  by  a  —  sign  may  be  I'emoved  if 
the  sign  of  each  term  enclosed  be  changed,  from  4-  to  — ,  andfi^om 
—  to  +. 

The  above  rules  apply  equally  to  the  removal  of  the  brackets, 
braces,  or  vinculum  (§  12). 

It  should  be  noticed  in,  the  case  of  the  latter  that  the  sign 
apparently  prefixed  to  the  first  term  underneath  is  in  reality 
prefixed  to  the  vinculum ;  thus,  -j-a  —  b  means  the  same  as 
4-  (a  —  b),  and  —a—b  the  same  as  —  (a  —  b). 

61.   1.   Remove  the  parentheses  from 

2  a -3  6 -(5  a -4  6)  + (4a- 6). 
By  the  rules  of  §  50,  the  expression  becomes 

2a-36-5a  +  46  +  4a-6  =  a. 


ADDITION   AND  SUBTRACTION  29 

Parentheses  sometimes  enclose  others  ;  in  this  case  they  may 
be  removed  in  succession  by  the  rules  of  §  50. 

Beginners  should  remove  one  at  a  time,  commencing  with  the 
innermost  -psiiv,  but  after  a  little  practice,  they  should  be  able 
to  remove  several  signs  of  aggregation  at  one  operation,  in  which 
case  they  should  commence  with  the  outermost  pair. 


2.    Simplify  4a;- S3aj-f(-2a;-a;-a)S. 

We  remove  the  vinculum  first,  then  the  parentheses,  and  finally  the 
braces. 

Thus,  4:X-{Sx-\-  {-2x-x-a)} 

=  4x-{Sx  +  {~2x-x  +  a)} 

=  4:X  —  {Sx  —  2x  —  gc-{-a] 

=  4:X  —  Sx  +  2x  +  x  —  a  =  4:X  —  a. 


EXERCISE  II 

Simplify  the  following  by  removing  the  signs  of  aggregation, 
and  then  uniting  similar  terms  : 

1.  9m  +  (— 4?M4-6n)  — (3  m  — n). 

2.  2x-Sy-l5x-\-y^-{-l-Sx~7y\. 
v^3.   a-6-2c-f- 2a-b-c -fa - 2 6 - c. 

4.   4:y^-23^-l-4.x^-7xy-{-5y^^  +  (Sa^-9xy). 


5.  3a^-5ab-{-^a'-{-2ab-9b^-7a'-6ab  +  b\ 

6.  5a-(7a-[9a  +  4]). 
^7.  7x-l-Sy-10x-lly}. 

♦8.  6  m?i  4-  5  —  ([  —  7  m7i  —  3]  —  f  —  5  mTi  — 11  J)- 


9.  8a2-9-(5a2-3a  +  2)  +  (6a2-4a-7). 

10.  2x-(Sy-\-5x-5x-y)-(-9y-^3x). 

»^11.  25- (-8- [-34-16-47]). 

12.  7a:-(5a;-[-12a;  +  6x-ll]). 

1/13.  2a-(-36  +  c- Ja-6S)-(3a  +  2c-[-r26  +  3c]). 


30  ALGEBRA 


14.  5m-[7m- J-3  m -4  wi  +  9| -6  m-8], 

15.  37 -[41-  513- (56-28  + 7) j]. 

16.  9  m  —  (3  ?2  +  j  4  m  —  [71  —  6  ??i]  \  —  [m  +  7  n]). 


17.    2a+[-6&- S3c  +  (-46-6c  +  a)|]. 


18.    7x-(-6x-l-5x-[-4.x-3x-2^l). 


19.  5  7i-[8?i-(37i  +  6)- J-6  7i  +  7  7i-5J]. 

20.  4a-[a-  5-7a-(8a-5a  +  3)-(-6a-2a-9)|]. 

21.  a;-S-ll2/-[2.T-(-42/-{-7aj-52/J-6aj-92/)]i. 


22.   3a-[5-(45-7c)-S2a-(36-5c)-66  +  cn.  ■ 
^23.    2a;-[-4a^-{5aj-(aj-7i»  +  6)j+(3ic-8a;-9)]. 

52.   Insertion  of  Parentheses. 

To  enclose  terms  in  parentheses,  we  take  the  converse  of  the 
rules  of  §  50. 

Any  number  of  terms  may  be  enclosed  in  parentheses  preceded 
by  a  -\-  sign,  without  changing  their  signs. 

Any  member  of  terms  may  be  enclosed  in  parentheses  preceded 
by  a  —  sign,  if  the  sign  of  each  tei^m  be  changed,  from  -{-  to  —, 
or  from  —  to  ■^. 

Ex.  Enclose  the  last  three  terms  of  a  —  b  +  c  —  d-\-e\n 
parentheses  preceded  by  a  —  sign. 

Result,  a  —  h  —  {—  c  -\-  d  —  e). 

EXERCISE  12 

In  each  of  the  following  expressions,  enclose  the  last  three 
terms  in  parentheses  preceded  by  a  —  sign  : 

1.   a  —  b  —  c-\-d.  \,5.4:Q^  —  y^  —  2yz  —  z\ 

9^2.   m3  +  2m_2-f  3m  +  4.  6.   a^ -{- b- -  (r  +  d\ 

3.   X'  -h  x^y  —  xy^  —  y\  7.   x^  —  2xy-\-y^-\-3x~4:  y. 

\/8.   ?2^-5w3-8?i2  +  6m  +  7« 


ADDITION  AND   SUBTRACTION  31 

9.  In  each  of  the  above  results,  enclose  the  last  two  terms  in 
parentheses  in  brackets  preceded  by  a  —  sign. 

53.  Addition  and  Subtraction  of  Terms  having  Literal  Coeffi- 
cients. 

To  add  two  or  more  terms  involving  the  same  power  of  a 
certain  letter,  with  literal,  or  numerical  and  literal,  coefficients, 
it  is  convenient  to  put  the  coefficient  of  this  letter  in  paren- 
theses. 

1.  Add  ax  and  2  x. 

Bj  §  40,  ax-\-2x  =  (a  +  2)x. 

2.  Add  (2  m  +  n)y  and  (m  —  3  n)y.        '  "^ 
{2  m  -h  n)y  +  (m  —  3  n)y  =  [(2  m  +  w)  +  (w  —  3  n)'\y 

=  (2  m  +  w  +  TO  -  3  w)?/(§  50)  =  (3  m  -  2  w)y. 
(The  pupil  should  endeavor  to  put  down  the  result  in  one  operation. ) 

3.  Subtract  (h  —  g)oi?  from  a^. 

By  §  48,  ax2  -  (h  -  'c)x^  =  [a-{b-  c)]x2 

=  (a-h  +  c)x''  (§  50). 

EXERCISE  13 

Add  the  following : 

1.   ax  and  bx.  4.   mx,  —nx,  and  —px. 

5.   aV  and  (ab  —  b^)x^. 
V6.  (3 a  +  4 b)n  and  (5 c-7 d)n. 


V2. 

mx^  and  —2a^. 

3. 

—  mny  and  —pqy. 

Subtract  the  following : 

^^. 

2  bx  from  3  ax. 

8. 

—  mny  from  aby. 

9.   —  7ixy  from  —  axy. 
10.    {p  +  q)x  from  mx. 
V 11.    (2  a  -  3  6)2/2  from  (5  a  -  4  b^f. 


32  ALGEBRA 


IV.    MULTIPLICATION  OF  ALGEBRAIC 
EXPRESSIONS 

54.  The  Rule  of  Signs. 

If  a  and  h  are  any  two  positive  numbers,  we  have  by  §  25, 

(+a)x(+6)=  +  a5,  (.^a)  x  (-6)  =  -a6, 

{—a)x{-\-h)  =  —  ab,  (_«,)  x  (— &)  =  +  a6. 

From  these  results  we  may  state  what  is  called  the  Rule  of 
Signs  in  multiplication,  as  follows  : 

V  TJie  product  of  two  terms  of  like  sign  is  positive;  the  product 
of  two  terms  of  unlike  sign  is  negative. 

55.  We  have  by  §  54, 

(—  a)  x{—b)x (—  c)  =  (ab)  x  (—  c) 

=  —  abc ;  (1) 

(—a)x{—b)x(—c)x(—d)=(—abc)x(—d),  by  (1), 

=  abcd;  etc. 

That  is,  the  product  of  three  negative  terms  is  negative ;  the 
product  of  four  negative  terms  is  positive ;  and  so  on. 
^  In  general,  the  product  of  any  number  of  terms  is  positive  or 
negative  according  as  the  number  of  negative  terms  is  even  or  odd. 

56.  The  Law  of  Exponents. 

Let  it  be  required  to  multiply  a^  by  al 

By  §  11,  a?=zaxax  a, 

and  a^  =  axa. 

Whence,  a^  x  a^  =  a  X  a  x  a  x  a  x  a  —  a\ 

We  will  now  consider  the  general  case. 

Let  it  be  required  to  multiply  a"*  by  a",  where  7n  and  n  are 
any  positive  integers. 


MULTIPLICATION   OF  ALGEBRAIC   EXPRESSIONS      33 

We  have  a™  =  a  x  a  X  •  •  •  to  m  factors, 

and  a''  =  axax  -"  to  n  factors. 

Then,  a"*  X  a''  =  a  x  a  x  •  ••  to  m  4-  71  factors  =  a*"+**. 

(The  Sign  of  Continuation^  •••,  is  read  '■'•and  so  on^) 

^^ence,  the  exponent  of  a  letter  in  the  product  is  equal  to  its 
exponent  in  the  multiplicand  plus  its  exponent  in  the  multiplier. 
This  is  called  the  Law  of  Exponents  for  Multiplication. 
A  similar  result  holds  for  the  product  of   three  or  more 
powers  of  the  same  letter. 

Thus,  a^xa''xa'  =  d^+^+^  =  a^. 

MULTIPLICATION  OF  MONOMIALS 

57.    1.   Let  it  be  required  to  multiply  7  a  by  —2  b. 
By  §31,  ^2b=(-2)xb. 

Then,  7  a  x  (- 2  &)=  7  a  x  (- 2)  x  6 

=  7  x(-2)x  a  X  6=-14a6.  (§54) 

In  the  above  solution,  we  assume  that  the  factors  of  a  product  can  be 
loritten  in  any  order. 

This  is  called  the  Commutative  Laic  for  Multiplication  ;  its  proof  for 
the  various  forms  of  number  will  be  found  in  §  453, 

2.   Eequired  the  product  of  —  2  a^b^,  6  ab^,  and  —  7  a^c. 
(-  2  a263)  X  6  a?>5  X  (-  7  a'^c) 

=  (-2)a253x  6a&5x(-7)«4c 

=  (— 2)  X  6  X  ( -  7)  X  a2  X  Of  X  a*  X  6^  X  6^  X  c 

=  84  aWc,  by  §§  55  and  56. 

We  then  have  the  following  rule  for  the  product  of  any 

number  of  monomials : 

^^  To  the  product  of  the  numerical  coefficients  (§§  30,  31,  55,  56) 
annex  the  letters ;  giving  to  each  an  exponent  equal  to  the  sum  of 
its  exponents  in  the  factors. 


34  ALGEBRA 

3.  Multiply  -5a%\)y  -S  ah\ 

(  -  5  a%)  X  (  -  8  ah^)  =  40  a^+i^i+a  ^  40  a*^*. 

4.  Find  the  product  of  4  n^,  —  3  71^,  and  2  n^. 

4  w2  X  (-  3  7z6)  X  2  w*  =  -  24  ^2+6+*.=  -  24  w". 

5.  Multiply  —  ic"*  by  7  a;^. 

6.  Multiply  6  (?)i  +  ny  by  7  (m  +  nf- 

6  (m  +  ny  X  7  (?n  +  w)^  =  42  (m  +  n)"^. 

c 

EXERCISE  14 

Multiply  the  following: 

1.  9  aj"^  by  4  ie2.  v"  9.   9(a +  &)' by  6(a  +  6f. 

2.  -8a^6by7a6^  10.    -  6  a*a^2/^  by  11  ccY^^ 
V  3.    11  aa;  by  -  3  61/.                     •^l.    -  2  a^-^sn  by  -  5  a^h'''. 

4.-7  0^2/^  by  —  9  aj^?/-       1^12.  14  a^^?/-"*  by  —  8  aj*?/*"- 
^  5.  15  6V  by  2  a^^a,  13^  4  ^^3^  _  7  ^5^  ^nd  -  3  m^ 

V  6.  -  a;'"?/"^'-  by  ^yh.  ^  14.  2  a^  6  6^  and  -  8  c«. 

7.  13(a;  — 2/)  by  —(x  —  yf.  15.  a^5',  6*c"»,  and  cV^. 

^8.  -  5  a^6V  by  -  12  a%^(?.         16.  -  5  a^y,  -  9y^V,  and-  ^'a; 
17.  2  a)^,  —  aj^,  6  x'^,  and  4  a;^. 
V 18.  -  3  a%  -  5  h%  -  2  c^a,  and  -  a^&V. 
/ 19.  3  m^TiV,  —  4  m^7iy,  —  5  m^y^y'^,  and  6  A'V. 

20.  a^m^?^  _  ^3p^r^  _  ^n^2^  and  -  ft^c. 

21.  m^n^^   —  2  ma;^,  3  mhf,  —  5  7iV,  and  —  4  n^?/^ 

MULTIPLICATION  OF  POLYNOMIALS  BY  MONOMIALS 

58.   In  §  40,  we  assumed  that  the  product  of  a  +  6  by  c  was 
ao,  H-  he. 


MULTIPLICATION   OF   ALGEBRAIC   EXPRESSIONS      35 

We  then  have  the  following  rule  for  the  product  of  a  poly- 
nomial by  a  monomial : 

V  Multiply  each  term  of  the  multiplicand  by  the  multiplier ^  and 
add  the  partial  products. 

Ex.     Multiply  23?-bx  +  l  by  -8a^. 
(2a;2_5x+7)x(-8a;3) 

=  (2:*:2)x(-8x3)4-(-5a;)x(-8a;3)  +  (7)x(-8x8) 

The  student  should  put  down  the  final  result  in  one  operation. 

EXERCISE  15 

Multiply  the  following : 
/I.   5a;-12  by  1  x.  5.   Sx^  by  Gar^  +  Sic-IT. 

2.  lOa^h  +  1  ah'  by  -Qab\  ^  6.  -4.a%^  by  3d'-2ab-4.h\ 
w/3.  a;*  —  4  x^y^-  +  4  2/^  by  -  xh/- 1  7.  7  x'^y^  —  8  x^y^  by  —  3  xhf. 
,4.   8m*-m2-3  by  5m^  ^8.   6a^-4a^-5a«  by  9a^ 

k  9.    —  m^n  +  8  ri^  —  3  m^  by  —  12  m^n^. 
^  10.   2  a^54  ^j  ^s  _Qa'b-\- 12  a^^  _  g  53^ 
HI.   97i2_6-2n^-5?i  +  7i^  by  -3n^ 
»^12.   5ar^-2i»27/_|_4ir?/2-32/^  by  lla^?/- 

4muLTIPLICATI0N  of  POLYNOMIALS  BY  POLYNOMIALS 

59.   Let  it  be  required  to  multiply  a-{-b  by  c  +  d. 

As  in  §  40,  we  multiply  a-{-b  by  c,  and  then  a-\-b  by  d, 
and  add  the  second  result  to  the  first;  that  is, 

(a  +  6)  (c  +  d)  =  (a  +  b)c  +  (a  +  &)d 

=  ac -\- be -^  ad -\-  bd. 

We  then  have  the  following  rule : 

Multiply  each  term  of  the  midtiplicand  by  each  term  of  the 
multiplier,  and  add  the  partial  pi'oducts. 


36  ALGEBRA 

60.     1.    Multiply  3a- 4 6  by  2 a -5b, 

In  accordance  with  the  rule,  we  multiply  3  a  — ib  by  2  a,  and  then  by 
—  5  &,  and  add  the  partial  products. 

A  convenient  arrangement  of  the  work  is  shown  below,  similar  terms 
being  in  the  same  vertical  column. 

3a  -4& 
2a  -5& 


6a^-   Sab 

-  15  a6  +  20  62 
6  a2  _  23  a&  +  20  &2 

The  work  may  be  verified  by  performing  the  example  with  the  multi- 
plicand and  multiplier  interchanged. 

2.   Multiply  ^a3(^  +  a^-8x^-2a^x  by  2x  +  a. 

It  is  convenient  to  arrange  the  multiplicand  and  multiplier  in  the  same 
order  of  powers  of  some  common  letter  (§  43),  and  write  the  partial 
products  in  the  same  order. 

Arranging  the  expressions  according  to  the  descending  powers  of  a,  we 

^^^®  a3-2a2x  +  4aa:2-8a;3 

a  +2x 


a*  -  2  a^x  +  4  a'^x^  -  8  ax^ 

2a^x-i  a2.r2  +  Sax^-  16  a;* 


a*  -16x* 

EXERCISE  16 

Multiply  the  following : 
VI.   5ir-7  by  3a;  +  2.  V4.    -lOxy  +  Shj-Sxy-A. 

2.  Sm-\-n  hj  8m-\-n,  5.  m^  —  m  —  3  by  m  +  3. 

3.  2a-S  hj  6a-7,  v  6.   a^ -a-12  hj  a-7. 

7.   4(a_6)_3by4(a-&)+3. 
^S.   x^-2xij-\-3y^hj  x-3y. 

9.   4m2+ 9n2-6mwby  37i4-2m. 
10.    ia_i6by  ia-i6. 
»/ 11.    a?  —  4  y  by  a;2  +  4  a;?/ 4- 16  2/". 


MULTIPLICATION   OF  ALGEBRAIC   EXPRESSIONS      37 

12.  a  +  6  +  c  by  a  —  5  —  c. 

13.  5  m^  +  3  m  —  4  by  6  r/i^  +  5  m^. 

14.  8  -  4  ?i  +  2  71^  -  ^3  by  2  +  7i. 

15.  2a'-8a-\-5hy  a--\-a-2. 

16.  6(m  +  ^)-  -  5(m  +  w)  +  1  by  7(m  +  n)-2. 

17.  2a^-3i»2-5aj-lby  3a;-5. 

18.  6ic  +  2a;2  +  8by  -4  +  a^2_3^_ 

19.  2  71^  +  m^  +  3  mn  by  2?^^  —  3  m7i  +  m^. 
120.  ^x'-ix  +  -^\  by  fa^  +  |. 

21.  4  a^  +  6  a  -  10  by  2  a-  -  3  a  +  5. 

v22.  9a7  +  2a^--5by4  +  3x'--7ic. 

23.  10  n^  -f-  3  n  -  4  by  9  71^  _  5  n  -  6. 

''24.  x^p+^y  —  x^y'i  by  x^p-'^ -{- y'^-\ 

25.  a'^  +  2  a-^d  +  2  ab'  +  6^  by  a^  -2ab-j-  h\ 

26.  m^-3m'''  +  97^2_27  ^^2,^81  by  m  +  3. 

\   27.  3(a4-^)'-2(a  +  ?>)+l  by  4(a  +  5)'-(a  +  6)4-5. 

28.  3  +  aV-7a-4a2by  a  +  a2-7. 

29.  8  m^  +  12  mhi  + 18  m^i^  +  27  r^^  by  2  ^^i^^i  -  3  mnK 

30.  4  a'"+^&2  _  3  ^4^„  i^y  ^m+25  _  2  «^,"-i. 

V31.  -a3-2a2  +  6a-5by  a--2a  +  10. 

32.  5i«*-6a^-4a^  +  2i«-3by  3a.'-2. 

V33.  4  m^  +  6  in'n  -  5  m^i^  -  3  n^  by  3  ^ti^  +  2  mw  -  wl 

^  35.  mx  +  m?/  —  nx  —  ny  by  ma?  —  my  +  nx  —  ny. 

36.  a»-3a2aj  +  3a£c2-a^by  a3  +  3a2a;  +  3aic2_^aj3. 

37.  a^-6a;2/  +  9/ by  ar^-9a^2^  +  27  0^2/2-27  2/3. 
^  38.  a"*  +  ^'^  —  e  by  a""  —  6"  +  e. 


38  ALGEBRA 

39.  2n3-3n2-n  +  4b.y  2n3-3?i2  +  n-4 

40.  5a^-7  +  2x'-Sxhj  -4:-^3x^-5x, 

41.  5a*  +  a-^-2a2-6a  +  3by2a--a-6. 

42.  Im'^-^m-l  by  \m-  +  \m-^' 

43.  a  +  3,  a  +  4,  and  a  — 5. 

44.  a;-6,  3aj-2,  and4a;  +  l. 

45.  m-{-2n,  w?  —  2  mn  +  4  w^,  and  m^  —  8  v?, 

46.  4cm  — 7,  5m  — S,  and  6 7?^  —  5. 
^47.  a;4-2,  .T  — 3,  i»-5,  and  i»  +  6. 

48.  a  +  26,  3a-4&,  and3a2-2a6-862. 

49.  2x  +  y,2x  —  y,4.x^  +  if,  and  16  x^  +  ^/^ 

50.  2  m  -\-  o  n,  2  m  —  3  n,  Z  m  -\-2  n,  and  3  m  —  2  w. 

51.  ^2  +  71+2,  w2-n  +  2,  andn*  +  3n2-4. 

752.    a-2,  a  +  3,  3a-l,  and  3  a^-2  a^- 19  a-6. 

61.  If  the  product  has  more  than  one  term  involving  the 
same  power  of  a  certain  letter,  with  literal,  or  numerical  and 
literal,  coefficients,  we  put  the  coefficient  of  this  letter  in  paren- 
theses, as  in  §  53. 

Ex.     Multiply  x^  —  ax  —  hx  +  ah  by  x  —  a. 

x'^  —  ax    —    hx  -{■  db 
x—a 


x^  —  ax?  —   hx'^  +  ahx 

—  ax^  +  a'^x  +  ahx  —  a'^h 

a:3  -^  (2  a  +  6)x2  +  (a2  +  2  ab^x  -  cfib 


As  in  §53,-2  ax"^  -  bx^  is  equivalent  to  -  (2  a  +  b)x^,  and  a^  +  2  abx 
to  (a2  +  2  ab)x. 

EXERCISE  17 

Multiply  the  following : 

yl.   a^  +  ax  +  bx  +  ab  hj  x-\-c. 


MULTIPLICATION   OF  ALGEBRAIC   EXPRESSIONS       39 

^2.  x^  —  mx -{- nx  —  mn  by  x—p. 

3.  x^  —  hx  —  cx-\-hc  by  x—  a. 

4.  x^  +  ax  —  hx  —  Z  ah  hj  X  -\-  h. 
y  b.  x^-{-ax  +  2hx-\-2  ab  hj  x  —  G. 

6.  x^ -{-px  —  5  qx  — 5pq  by  x  —  r. 

7.  0/'^  —  3ax  —  bx-\-3ab  by  cc  +  2  a. 

8.  x^  —  4:  mx  -\-nx  —  4:  mn  by  x-\-Sn. 
^d.  x^-^3ax-2bx  +  6ab  hj  x-4:C. 

10.  (a-6>-3a&  by  2a;-(a-6). 

11.  x^^  —  B  ax""  +  4  bx''  —  2ab  by  x"  +  c. 

V/  12.    (2  a  -  l)a;2  +  (a  +  2)x  -  (a  +  3)  by  (a  -  2)x  -  a. 

yC»62.   Ex.     Simplify  (a-2  xY-2(S  a-i-x)(a-x). 

To  simplify  the  expression,  we  first  multiply  a  —  2x'by  itself  (§  11)  ;  we 
then  find  the  product  of  2,  3  a  +  x,  and  a  —  x,  and  subtract  the  second 
result  from  the  first. 

a  —2x  Sa  +  X 

a  —2x  a  —  X 


2  ax  3  o2  4-     Ota; 

2 ax  +  4 jc2  -Sax  - 


a^-4:ax  +  ix^  Za^-2ax-    x^ 

2 


6a'^-4:ax-2x^ 
Subtracting  the  second  result  from  the  first,  we  have 

a"^  -  iax  +  ix"^  -  Q  a^  -h  4:ax  +  2x^  =  -  6  a'^  -^  6x^, 

EXERCISE  18 
Simplify  the  following : 

1.    (3a  +  5)(2a-8)4-(4a-7)(a-h6). 
v2.   (Sx-{-2)(4.x  +  S)-(Sx-2)(4.x-S), 
v/3.    (a-2 x)(b  +  3y)  +  (a-{-2 x)(b-3y). 


40  ALGEBRA 


4. 

5. 

6. 

^   7. 

8. 

9. 

10. 

11. 

12. 

13. 

14. 

15. 

16. 

^17. 

^  18. 

*19. 

20. 

•  21. 

22. 

23. 

24. 


3m  +  iy{Sm-lf. 

2a  +  3  6)2-4(a-6)(a  +  5  6). 

3x-{5y-j-2z)-][_3x~(5y-2z)-]. 

711  +  2  n  — (2  m  —  n)]  [2  m  +  n  —  (m  —  2  n)]. 

a-{-b)-{-c'-(a-b-cy. 

a  +  2)(a  +  3)(a  _  4)  +  (a  -  2)(a  -  3)(a  +  4). 

i^-iy  +  i^y- 

2x'+(3x-l)(4.x-i-5)^l5x'-(4:X  +  3)(x-2)']. 
a  +  2b-c-3  ay. 

a  -  2)(a  +  3)  -  (a  -  3)(a  +  4)  -  (a  -  4)(a  +  5). 
a;  +  2)(2  X  - 1)(3  05  -  4)  -  (oj  -  2)(2  a.' + 1)(3  x  +  4). 

^  -  (2/  -  ^)]  [2/  -  (^  -  2=)]  [2=  -  (»  -2/)]- 
a  -  5)(a3  4_  6•3^)  |-(^(^  _^  ^)  _^  ^^2-]^ 

a-^b-2cy-(b-\-G-2ay+(c  +  a-2by. 
x+y+zy+(x-y-zy+(-x+y-zy+{-x-y+zy. 

2x-i-iy-\-(2x-iy. 

a-\-b  +  c)(ab  -\-bc  +  ca)  —  (a  +  b)(b  +  c)(c  +  a). 

a  +  2&)2-2(a  +  25)(2a  +  &)  +  (2a  +  Z')'. 

a;  +  2/  +  2;)=^  -  3  (2/  +  2;)  (2;  +  a^)  (oj  +  2/) . 

a-\-by  +  3(a-\-by(a-b)-\-3(a  +  b)(a-by  +  (a-by 

DEFINITIONS 


63.  A  monomial  is  said  to  be  rational  and  integral  when  it 
is  either  a  number  expressed  in  Arabic  numerals,  or  a  single 
letter  with  unity  for  its  exponent,  or  the  product  of  two  or 
more  such  numbers  or  letters. 

Thus,  3  a^W,  being  equivalent  to  3  •  a  •  a  •  5  •  6  •  &,  is  rational 
and  integral. 


MULTIPLICATION  OF   ALGEBRAIC   EXPRESSIONS      41 

A  polynomial  is  said  to  be  rational  and  integral  when  each 
terra  is  rational  and  integral ;  as  2  ic^  —  -a6  +  cl 

64.  If  a  term  has  a  literal  portion  which  consists  of  a  single 
letter  with  unity  for  its  exponent,  the  term  is  said  to  be  of  the 
fifbt  degree. 

Thus,  2  a  is  of  the  first  degree. 

The  degree  of  any  rational  and  integral  monomial  (§  63)  is 
the  number  of  terms  of  the  first  degree  which  are  multiplied 
together  to  form  its  literal  portion. 

Thus,  5  a&  is  of  the  second  degree ;  3  a^W,  being  equivalent 
to  3  •  (X  •  a  •  &  •  6  •  &,  is  of  the  fifth  degree ;  etc. 

The  degree  of  a  rational  and  integral  monomial  equals  the 
sum  of  the  exponents  of  the  letters  involved  in  it. 

Thus,  a6V  is  of  the  eighth  degree. 

The  degree  of  a  rational  and  integral  polynomial  is  the 
degree  of  its  term  of  highest  degree. 

Thus,  2  a^6  —  3  c  +  d^  is  of  the  third  degree. 

65.  Homogeneity. 

Homogeneous  terms  are  terms  of  the  same  degree. 
Thus,  a^^  3  bh,  and  —  5  x^y'^  are  homogeneous  terms. 
A  polynomial  is  said  to  be  homogeneous  when  its  terms  are 
homogeneous ;  as  a^  +  3  &^c—  4  xyz. 

66.  If  the  multiplicand  and  multiplier  are  homogeneous,  the 
product  will  also  be  homogeneous,  and  its  degree  equal  to  the 
sum  of  the  degrees  of  the  multiplicand  and  multiplier. 

The  examples  in  §  60  are  instances  of  the  above  law ;  thus, 
in  Ex.  2,  the  multiplicand,  multiplier,  are  homogeneous,  and  of 
the  third,  first,  and  fourth  degrees,  respectively. 

The  student  should  always,  when  possible,  apply  the  prin- 
ciples of  homogeneity  to  test  the  accuracy  of  algebraic  work. 

Thus,  if  two  homogeneous  expressions  be  multiplied  together, 
and  the  product  obtained  is  not  homogeneous,  it  is  evident  that 
the  work  is  not  correct. 


42  ALGEBRA 


V.  DIVISION    OF    ALGEBRAIC    EXPRESSIONS 

67.  We  define  Division,  in  Algebra,  as  the  process  of  finding 
one  of  two  numbers,  when  their  product  and  the  other  number 
are  given. 

The  Dividend  is  the  product  of  the  numbers. 
The  Divisor  is  the  given  number. 
The  Quotient  is  the  required  number. 

68.  The  Rule  of  Signs. 

Since  the  dividend  is  the  product  of  the  divisor  and  quotient, 
the  equations  of  §  54  may  be  written  as  follows : 

±^=+b,   =^  =  +  h,   =^  =  -6,   and  +^  =  -6. 

+  a  —a  -\-a  —a 

Froijd  these  results,  we  may  state  the  Rule  of  Signs  in  divi- 
sion, as  follows : 

Tlie  quotient  of  two  terms  of  like  sign  is  positive;  the  quotient 
of  two  terms  of  unlike  sign  is  negative. 

69.  Let  -  =  aj.  (1) 

0 

Then,  since  the  dividend  is  the  product  of  the  divisor  and 
quotient,  we  have  a  =  bx 

Multiply  each  of  these  equals  by  c  (Ax.  7,  §  9), 

ac  =  hex. 

Eegarding  ac  as  the  dividend,  be  as  the  divisor,  and  x  as  the 
quotient,  this  may  be  written 


be 

(2) 

From  (1)  and  (2), 

f5  =  ?.    (Ax.  4,  §9) 

be      b 

(3) 

That  is,  a  factor  common  to  the  dividend  and  divisor  can  be 
removed,  or  cayicelled. 


DIVISION  OF   ALGEBRAIC   EXPRESSIONS  43 

70.  The  Law  of  Exponents  for  Division. 

Let  it  be  required  to  divide  a^  by  al  ^ 

By  §  11, 

w  a  X  a 

Cancelling  the  common  factor  «  x  a  (§  69),  we  have 

—  =  a  xaxa  =  a^. 
a" 

We  will  now  consider  the  general  case. 

Let  it  be  required  to  divide  a"*  by  a",  where  m  and  n  are  any 
positive  integers  such  that  m  is  greater  than  n. 

■VKT    X.  a"*  _  ax  axa  X  "-  to  m  factors 

a'*      a  X  a  X  a  X  •  •  •  to  ?i  factors 

Cancelling  the  common  factor  axaxax-"ton  factors, 

—  =  axaxax  •••torn  —  n  factors  =  a"""". 

Hence,  the  exponent  of  a  letter  in  the  quotient  is  equal  to  its 
exponent  in  the  dividend,  minus  its  exponent  in  the  divisor. 
This  is  called  the  Laio  of  Exponents  for  Division. 

DIVISION  OF  MONOMIALS 

71.  1.  Let  it  be  required  to  divide  —  14a"6  by  7a^. 

Cancelling  the  common  factors  7  and  a^  (§  69),  we  have 

Then  to  find  the  quotient  of  two  monomials : 
To  the  quotient  of  the  numerical  coefficients  annex  the  letter Sj 
giving  to  each  an  exponent  equal  to  its  exponent  in  the  dividend 
minus  its  exponent  in  the  divisor,  and  omitting  any  letter  having 
the  same  exponent  in  the  dividend  and  divisor. 


44  ALGEBRA 

2.   Divide  54  a'hh^  by  -  9  a'h\ 
54  a563c2 


-  9  a*63 
3.    Divide  —2x^'^y''z''  by  —  a;'"?/V, 


=  -  6  a5-4c2  =  -  6  ac2. 


4.   Divide  35  (a  -  6)^  by  7  (a  -  b)\ 


EXERCISE  19 

Divide  the  following : 

1.  30by-5.  4.   -64by8.  7.  -^by^. 

2.  -  42  by  6.  5.   - 135  by  -  9.       8.   21  a^'  by  3  al     . 

V  3.  -  48  by  -  4.  ~  6.  176  by  - 11.    9.  -63mV  by  7mV. 

10.  6  a^2/'°  by  -  x^'-  -  18.  -  28  a'b'c'  by  2  6V. 

'•11.  9  (a -6/ by  3  (a -6)2.    19.  a^+^b^+^  hj  -  ab\ 

12.  xy'z^  by  -'rc^/^.        V  20.  -  55  afyh'''  by  -  11 2/V. 

13.  - 13  ?/!%•'  by  -  13  ?/i5?i^       21.  -  70  a^6V  by  14  ab^c\ 

14.  45(a;  +  2/y  by -5(a;  +  ?/y.   22.   -  32  o.'^^^/'"^'- by  -  8  a;  V- 

15.  72a;yby6a;y.  23.   -  96  x'^'-Y  ^J  ^^  ^""^Y- 
,  16.  -  40  a'b'c'  by  -  8  be.  24.   52  a^^^e^ia  by  _  4  a'b(f. 

17.  90a"a:«by  9aV.  25.   132  a^y^;!^  by  12  a^y^^. 

Find  the  numerical  value  when  a  =  2,  &  =  — 4,  c  =  5,  and 
(«  =  -3of: 

2Q    10 ab     Sac  28    '^«  +  14&  — 12c 

cc?         6d  '  '   13  a  -  9  6  +  17  c' 

rtty    2  ft  —  6  _  ct  4-  4  &  29      a  — &    ,     6  —  c        c  —  (f 


5d     3c4-ci  a  +  36     &  +  5c     c  +  4(^ 


DIVISION   OF   ALGEBRAIC   EXPRESSIONS  45 

DIVISION  OF  POLYNOMIALS  BY  MONOMLAXS 

72.   We  have,  (a  +  h)c  =  ac-\-  he. 

Since  the  dividend  is  the  product  of  the  divisor  and  quotient 
(§  67),  we  may  regard  ac  +  he  as  the  dividend,  c  as  the  divisor, 
and  a  4-  6  as  the  quotient. 

Whence,  ^^±i2  =  a4-&. 

c 

Hence,  to  divide  a  polynomial  by  a  monomial,  we  divide  each 
term  of  the  dividend  hy  the  divisor,  and  add  the  results, 

Ex.   Divide  9  a^h^  -^^"0  + 12  a^hc^  by  -  3  a\ 

—  3  a2 

EXERCISE  20 
Divide  the  following : 

V  1.   25a«-15a«  +  40a^  by  5a^ 

">    2.    —  24  mV  +  33  mn^  by  —  3  mn^. 

V  3.   S6  xhjz^- 9  xYz^- 27  a^ifz'hy  9  a^y. 
V '  4.   54  a'b'c^  -  60  a^b^c^  by  6  ab'c'. 

K    5.  -  22  x^V  +  30  xY  +  26  xy  by  -  2  xy. 

6.  70ni3-56w"-63w9  +  49?/by7  7i^ 

*  7.  66  a;'^?/^  +  77  xy^z  —  55  xyz^  by  —  11  xyz. 

8.  36ai^  +  28ai2_4^9_20a6by4a«. 

V  9.  ic^+?2/"+2  —  a;«+y  by  a^/. 

10.   14  m^n2  -  28  m^7i^  +  28  m^w^  - 14  wiV  by  - 14  mV. 
•  11.   32a;i5  +  24.'»^3_43^n_40^9]^y  _8aj8^ 

12.   84  oi^y^z^  - 108  icy^s  _  48  xYz'  by  12  xyz^ 
'^IS.   a^pfe'c^'- -  a^&^c^'- -  a^^6^^c^'' by  -a^¥c-'\ 
"^  14.   30  a^^mhi^  —  60  a^m^n^  —  45  ci^ri^  by  — 15  a^n^. 


46  ALGEBRA 

DIVISION  OF  POLYNOMIALS  BY  POLYNOMIALS 

73.   Let  it  be  required  to  divide  12  + 10  oj^  - 11  a;  -  21  a;^  ^^ 

Arranging  each  expression  according  to  the  descending 
powers  of  x  (§  43),  we  are  to  find  an  expression  which,  when 
multiplied  by  the  divisor,  2  a^  —  3x  —  4:,  will  produce  the 
dividend,  10  x^  -  21  cc^  -  11  a;  +  12. 

It  is  evident  that  the  term  containing  the  highest  power  of 
X  in  the  product  is  the  product  of  the  terms  containing  the 
highest  powers  of  x  in  the  multiplicand  and  multiplier. 

Therefore,  10  x^  is  the  product  of  2  x^  and  the  term  contain- 
ing the  highest  power  of  x  in  the  quotient. 

Whence,  the  term  containing  the  highest  power  of  x  in  the 
quotient  is  10  x^  divided^  by  2  x",  or  5  x. 

Multiplying  the  divisor  by  5  x,  we  have  the  product 
10  0^—15x^  —  20  X',  which,  when  subtracted  from  the  divi- 
dend, leaves  the  remainder  —6x^-\-9x-\-12. 

This  remainder  must  be  the  product  of  the  divisor  by  the 
rest  of  the  quotient ;  therefore,  to  obtain  the  next  term  of  the 
quotient,  we  regard  — 6a^-f-9aj4-12  as  a  new  dividend. 

Dividing  tlie  term  containing  the  highest  power  of  x,  —  6x% 
by  the  term  containing  the  highest  power  of  x  in  the  divisor, 
2  a^,  we  obtain  —  3  as  the  second  term  of  the  quotient. 

Multiplying  the  divisor  by  —  3,  we  have  the  product 
—  6  a.'^  +  9  a;  + 12 ;  which,  when  subtracted  from  the  second 
dividend,  leaves  no  remainder. 

Hence,  5  a;  —  3  is  the  required  quotient. 

2  a^  —  3  a;  —  4,  Divisor.  ■  - 


10aj3-21aj2-lla.'-|-12 
10a^-15a;2-20'a; 


5  a;  —  3,  Quotient. 


-  6a^-f    9a;  +  12 

-  6x^-{-    9a; +  12 


The  example  might  have  been  solved  by  arranging  the  dividend  and 
divisor  according  to  the  ascending  powers  of  x. 

From  the  above  example,  we  derive  the  following  rule. 


DIVISION   OF   ALGEBRAIC   EXPRESSIONS  47 

Arrange  the  dividend  and  divisor  in  the  same  order  of  powers 
of  some  common  letter. 

Divide  the  first  term  of  the  dividend  by  the  first  term  of  the 
divisor,  and  ivrite  the  result  as  the  first  term  of  the  quotient. 

Multiply  the  ivhole  divisor  by  the  first  term  of  the  quotient,  and 
subtract  the  product  from  the  dividend. 

If  there  be  a  remainder,  regard  it  as  a  new  dividend,  arid, 
proceed  as  before;  arranging  the  remaiyider  in  the  same  order 
of  powers  as  the  divideiid  and  divisor. 

1.   Divide  ^ab"" +  a^-^W-^a-b  by  ^W-{-d'-2ab. 
Arranging  according  to  the  descending  powers  of  «, 

^3  _  5  ^25  _,.  9  ^52  _  9  ^,3  f.  05?  _  2  a&  4-  3  62 


Sb 


3  a^b  +  6  a62 
Sa^b  +  6a&2_9^3 


In  the  above  example,  the  last  term  of  the  second  dividend  is  omitted, 
as  it  is  merely  a  repetition  of  the  term  directly  above. 

The  work  may  be  verified  by  multiplying  the  quotient  by  the  divisor, 
which  should  of  course  give  the  dividend. 

2.   Divide  4.  +  9x^-2Sa^  by  -3x^  +  2+4:X. 
Arranging  according  to  the  ascending  powers  of  x, 

2  4-  4  X  -  3  x2 


4  -  28  a;2  +    0  x4 
4+    8  a;  -    6x2 


2  -  4  X  -  3  ic2 


-  8  X  -  22  X-  +    9  X* 

-  8  X  -  16  x2  4- 12  x3 


6  x2  -  12  x3  +  9  X* 
6  x2  -  12  x3  4-  9  x* 


EXERCISE  21 
Divide  the  following : 

1.   6a^  +  29a4-35  by  2a-{-5. 
^2.   S0x'-53x-\-S  by  6x-l. 

3.   32a.-2  +  28i»2/-15/  by  4.x-{-5y. 
»^4.   10a3  +  33a2_52a  +  9  by  5a-l, 


48  ALGEBRA 

5.  a^-Sb^  by  a-2b. 

■  6.  2Sn^  +  S4:n-12  by  -7n-\-2. 

7.  64a^  +  27/  by  4.x-\-3y. 

V  8.  6(x-yy-7(x-y)-20  by  3(a;-2/)+4. 

9.  25m^n^  —  S6  by  6  +  5mn. 

via  -25«3^  +  12a;^  +  12a.'2/  by  Sx'-Axy. 

11.  18m3-17 maj2-6a^  by  3m4-2ic. 

12.  27v'-6-\-5n^-19n  hj  -Sn-{-5n^-3. 

13.  12-{-13x^-19x-12x' hy  -Sx^-4.-\-x, 

14.  x'^  —  y-  —  2yz  —  z^hjx  —  y  —  z. 

15.  8(a  +  &)^-c^  by  2(a  +  6)-c. 

16.  8  m'^/i  —  24  m^n^  + 12  ?n^  —  31  m^7i^  by  6m^  — 8n^  — 5mn. 

17.  8a  +  9a^-l-16a2  by  l-3a2-4a. 

18.  |m^  — ^m  — I  by  |m  +  ^« 

vl9.  2a^-10-6a^-{-x^-{-llx  hj  2  +  x'^-x. 

20.  a;^-81  by  a;3-3  a;2+ 9  a;-27. 

v'21.  a^-256  6^  by  a -4  6. 

22.  m^  —  7  mV  +  w^  by  m^  +  3  mw  +  n^. 

23.  81a;^-l  by  l+3aj. 

24.  6-a3  +  6a^-8a-23a2  by  2a  +  3. 

^25.  {x-^yf-9{x-\-yy  +  27(x-\-y)-27  by  (x  +  y)-3. 

26.  -36ic2-l+4a;^-12aj  by  -2x^-\-l  +  6x. 

27.  10a-a2-25  +  16a*  by  5  +  4a2-a. 

28.  ^0^3^21  by  |aj  +  |/ 

^29.   37i*-lln3-25w2-137i-2  by  37i2  +  47i  +  l. 
30.   2a;y  +  2/^  +  9a;^  by  -  2  a;^/ +  3  a;^  +  2/^. 
.  31.   73x  +  37a^-35-h20x^-15x^  hj  -5-^4:x''  +  9x. 


DIVISION  OF  ALGEBRAIC  EXPRESSIONS  49 

32.  243n^  +  l  by  3w  +  l. 

33.  x'-^16i^  +  96a^  +  256x-{-256  by  (a;  +  4)2. 

34.  _60w^  +  127n2  +  214ri-336  by  - 12  n^ -\- 11  n  +  5Q. 

35.  -324-a'  by  8a  +  a4  +  16  +  2a3  +  4(x2. 

v36.  a^'^+^ft-"  +  2  a2'«+352«+i  _^  ^2m+552n+2  j^^y  a"*6«-i  +  a'^+S^n 

37.  2V^'-tV«'^+tV«^'-6V?>'  by  i«-i&. 

38.  5aj3  +  24-33a;2  +  l0a;  +  3a;^  by  -4  +  3a;. 

39.  aj^  +  37a^-70a;  +  50  by  a^-2aj  +  10. 
''  40.  0^+^  +  8  a;^-^  by  aj2«»+i  _  2  a^"*  +  4  a^'»-i. 

/41.  {Sd'  +  5a-2)(2d'-a-6)  by  (3  a-l)(2a  +  3). 

42.  63aj^  +  114ar^  +  49a^-16a;-20  by  dx'  +  Qx-Q. 

v^43.  a^+i^^  -  a&^^+'  by  a^+'^b^-ab'^+K 

,44.  a'^_65_5(^45_^5^54_j_-L()^3^2_jL0a263 
by  a^-63-3a26  +  3a6l 

45.  _67i^-25n4 4-77^^  +  817^2  +  371-28 

by   _27i3-5n2  +  87i  +  7. 

46.  23a^-5aj4-12  +  12a^  +  8a;-14a^  by  a;-2  +  3a^. 

47.  15a3-4a^-15  +  8a«-5a-2a^+3a2  by  -a-^-Aw'-S. 

48.  52aj3  +  64  +  18a;^-200.'»2  +  a^  by  6a;2- 8 +a;3_i2a;. 
,  49.  a-"^  —  62"  _  2  6"c^  —  c^p  by  a"*  —  6"  —  c^. 

50.  a^-6aV  +  9a27i^-47i«  by  a^ -  2  a^n  - an^ -\- 2  n^. 

51.  3aj^-7a;^-llic^  +  5a^  +  7a;2  +  5a;-2  by  3a^-a?-2. 

52.  57i3+6n-ri^-8  +  67i6+467i2-38n* 

by  2-57i2  +  3n3-47io 

53.  ^m*-2m^  +  lm^-j\  hj  ^m^-^m-l, 

54.  9»2_25  2/2_40  2/2_l6.;22  by  3aj4.52/  +  42!. 

^  55.  90  71*  - 143  n^  - 102  n^  +  131  ti  +  60  by  (2  tj  -  3)(5  n  +  4). 

^   56.  «*"»  +  a;2m^4n  _|.  2^n  by  x^""  —  x"'y^'' +  y^. 


50  ALGEBRA 

^"57.   4.x^-21  xy  +  21  xY  -  4  2/6  by  (x-y)(x-2  y)(2x-\-  y). 

V  58.   a^  +  32  +  10a(a3  +  8)+40a2(a  +  2)  by  (a-2)2  +  8a. 

74.  By  §  66,  if  the  dividend  and  divisor  are  homogeneous, 
the  quotient  will  be  homogeneous,  and  its  degree  equal  to  the 
degree  of  the  dividend  minus  the  degree  of  the  divisor. 

75.  The  operation  of  division  is  often  facilitated  by  the  use 
of  parentheses. 

JEx.     Divide  x^-}-(a-{-b—c)x^+(ab—bc—ca)x—abc  by  x+a. 

ic^  +  (a  4-  &  —  c)x2  +  (a6  —  be  —  cd)x  —  ahc  I  x_  Ya       

x^+ [^  I  »^  4-  C&  -  c)x  -  be 

(6  -  c)x2 

(b  —  c)x'^  4-  {ab  —  ca)x 

—  bcx 

—  bcx  —  abc 

EXERCISE  22 

Divide  the  following : 

^  1.   a^  ■\-  (a  —  b  —  g)x^  +  (—  ab  -i-bc  —  ca)x  +  abc 
by  x^  +  (a  —  b)x  —  ab. 

2.  a^  +  (a  +  6  —  c)a;^  +  (a6  —  &c  —  ca)x  —  a6c  by  ic  —  c. 

3.  :i?—{a-\-b-{-c)x^-\-(ab-\-bc-{-ca)x—abG  by  a;^— (6  +  c)a;+6c. 

4.  a^  -  (a  -  2  6  -  3  c)^?^  -f(_2a6  +  6  6c-3  ca)x  -  6  a6c 

by  ic^  —  (a  —  3  c)a;  —  3  ac. 

5.  a:3+  (3  a  +  &4-2 c)a^+  (3 a&+2  6c4-6 ca)x+^  abc  by  a;+3 a. 
.       6.    a{a-b)x?+{-ab-\-b'^-\-bc)x-c{b+c)  by  (a-6)fl?+c. 

V  7.   m(m  +  w)flj^  —  (m^  +  n^)a;  +  n{m  —  n)  by  ma;  —  n. 

8.  a:;^  —  (m  —  2  n)a;  —  2  m^  -f- 11  mw  —  15  w^  by  a;  +  m  —  3  w. 

9.  (2  m2  +  10  mn)x^  +  (8  m^  -  9  mn  -  15  n'^)x  -  (12  mw  -  9  ii^ 

by  2  ma;  —  3  n. 

V  10.   a;3-(3a4-26-4c)a;''^+(6a6-86c+12ca)a;~24a5cbya;-26. 


INTEGRAL  LINEAR  EQUATIONS  51 


VI.  INTEGRAL  LINEAR  EQUATIONS 

76.  Any  term  of  either  member  of  an  equation  is  called  a 
term  of  the  equation. 

77.  A  Numerical  Equation  is  one  in  which  all  the  known 
numbers  are  represented  by  Arabic  numerals ;  as, 

2x  —  l  —  x-\-Q>. 

An  Integral  Equation  is  one  each  of  whose  members  is  a 
rational  and  integral  expression  (§  63)  ;  as, 

78.  An  Identical  Equation,  or  Identity,  is  one  whose  members 
are  equal,  whatever  values  are  given  to  the  letters  involved ; 
as  (a  -\-h)  (a  —  h)  =  o?  —  61 

The  sign  =,  read  "^s  identically  equal  to,''''  is  frequently  used  in  place 
of  the  sign  of  equality  in  an  identity. 

79.  An  equation  is  said  to  be  satisfied  by  a  set  of  values  of 
certain  letters  involved  in  it  when,  on  substituting  the  value 
of  each  letter  in  place  of  the .  letter  wherever  it  occurs,  the 
equation  becomes  identical. 

Thus,  the  equation  x  —  y  =  5  is  satisfied  by  the  set  of  values 
x  =  S,  y  =  S;  for,  on  substituting  8  for  x,  and  3  for  y,  the  equar 
tion  becomes  8  —  3  =  5,  or  5  =  5;  which  is  identical. 

80.  An  Equation  of  Condition  is  an  equation  involving  one 
or  more  letters,  called  Unknown  Numbers,  which  is  satisfied 
only  by  particular  values  of  these  letters. 

Thus,  the  equation  a?  +  2  =  5  is  not  satisfied  by  every  value 
of  X,  but  only  by  the  particular  value  x  —  S. 

An  equation  of  condition  is  usually  called  an  equation. 

Any  letter  in  an  equation  of  condition  may  represent  an 
unknown  number ;  but  it  is  usual  to  represent  unknown  num- 
bers by  the  last  letters  of  the  alphabet. 


52  ALGEBRA 

81.  If  an  equation  contains  but  one  unknown  number,  any 
value  of  the  unknown  number  which  satisfies  the  equation  is 
called  a  Root  of  the  equation. 

Thus,  3  is  a  root  of  the  equation  x-\-2  =  5. 
To  solve  an  equation  is  to  find  its  roots. 

82.  If  a  rationaland  integral  monomial  (§  63)  involves  a 
certain  letter,  its  degree  with  respect  to  it  is  denoted  by  its 
exponent. 

If  it  involves  two  letters,  its  degree  with  respect  to  them  is 
denoted  by  the  sum  of  their  exponents ;  etc. 

Thus,  2  aWx^y^  is  of  the  second  degree  with  respect  to  x,  and 
of  the  fifth  with  respect  to  x  and  y, 

83.  If  an  integral  equation  (§  77)  contains  one  or  more  un- 
known numbers,  the  degree  of  the  equation  is  the  degree  of  its 
term  of  highest  degree. 

Thus,  if  X  and  y  represent  unknown  numbers, 

ax—by  =  c  is  an  equation  of  the  first  degree ; 
a^-\-4:X=  —  2,  an  equation  of  the  second  degree; 
2x^—3xy^  =  5,  an  equation  of  the  third  degree ;  etc. 

A  Linear,  or  Simple,  Equation  is  an  equation  of  the  first  degree. 

PRINCIPLES  USED  IN  SOLVING  INTEGRAL  EQUATIONS 

84.  Since  the  members  of  an  equation  are  equal  numbers, 
we  may  write  the  last  four  axioms  of  §  9  as  follows : 

1.  TJie  same  number,  or  equal  numbers,  may  be  added  to  both 
members  of  an  equation  ivithout  destroying  the  equality. 

2.  Hie  same  number,  or  equal  numbers,  may  be  subtracted 
from  both  members  of  an  equation  ivithout  destroyiyig  the  equality. 

3.  Both  members  of  an  equation  may  be  multiplied  by  the  same 
number,  or  equal  numbers,  without  destroying  the  equality. 

4.  Both  members  of  an  equation  may  be  divided  by  the  same 
number,  or  equal  numbers,  without  destroying  the  equality. 


INTEGRAL   LINEAR  EQUATIONS  53 

85.  Transposing  Terms.  , 

Consider  the  equation  x-\-a  —  h  =  c. 

Adding  —  a  and  +  6  to  both  members  (§  84,  1),  we  have 

x=c  —  a-\-h. 

In  this  case,  the  terms  +  a  and  —  b  are  said  to  be  transposed 
from  the  first  member  to  the  second. 

Hence,  any  term  may  he  transposed  from  one  member  of  an 
equation  to  the  other  by  changing  its  sign. 

86.  It  follows  from  §  85  that 

If  the  same  term  occurs  in  both  members  of  an  equation  affected 
with  the  same  sign,  it  may  be  cancelled. 

87.  Consider  the  equation 

a  —  x  =  b  —  c.  (1) 

Multiplying  each  term  by  —  1  (§  84),  we  have 

x  —  a  =  c  —  b', 

which  is  the  same  as  equation  (1)  with  the  sign  of  every  term 
changed. 

Hence,  the  signs  of  all  the  terms  of  an  equation  may  be  changed, 
without  destroying  the  equality. 

88.  Clearing  of  Fractions. 

Consider  the  equation 

2  5     5        9 

-X =  -x 

3  4     6        8 

Multiplying  each  term  by  24,  the  lowest  common  multiple  of 
the  denominators  (Ax.  7,  §  9),  we  have 

16a;-30  =  20a;-27, 
where  the  denominators  have  been  removed. 

Kemoving  the  fractions  from  an  equation  by  multiplication 
is  called  clearing  the  equation  of  fractions. 


54  ALGEBRA 

SOLUTION  OF  INTEGRAL  LINEAR,  EQUATIONS 

89.  To  solve  an  equation  involving  one  unknown  number, 
we  put  it  into  a  succession  of  forms,  which  finally  leads  to  the 
value  of  the  root. 

This  process  is  called  transforming  the  equation. 
Every  transformation  is  effected  by  means  of  the  principles 
of  §§  84  to  88,  inclusive. 

90.  Examples. 

1.  Solve  the  equation 

5a;  — 7  =  3a;  +  l. 
Transposing  3ic  to  the  first  member,  and  —  7  to  the  second  (§  85),  we 
^^^^  5x-3x  =  7  +  L 

Uniting  similar  terms,  2  sc  =  8. 

Dividing  both  members  by  2  (§  84,  4), 

To  verify  the  result,  put  cc  =  4  in  the  given  equation. 

Thus,  20  -  7  =  12  +  1 ;  which  is  identical. 

2.  Solve  the  equation 

6        3~6        4 

Clearing  of  fractions  by  multiplying  each  term  by  60,  the  L.  C.  M.  of 
6,  3,  5,  and  4,  we  have 

70  cc  -  100  =  36  X  -  15. 

Transposing  36  x  to  the  first  member,  and  — 100  to  the  second, 

70a;-36x  =  100-15. 

Uniting  terms,  34  a;  =  85. 

Divided  by  34,  x  =  —  =  -. 

34     2 

3.  Solve  the  equation 

(5 -3  a;) (3 +  4 ic)  =  62 -  (7-3 a;)(l -"4a;). 


INTEGRAL  LINEAR   EQUATIONS  65 

Expanding,       15  +  11  x  -  12  x^  =  62  -  (7  -  31  x  +  12  x2). 
Or,  15  +  11  X  -  12  x2  =  62  -  7  +  31  X  -  12  x\ 

Cancelling  the  —  12  x^  terms  (§  86),  and  transposing, 

llx-31x  =  62-7-15. 
Uniting  terms,  —  20  x  =  40. 

Dividing  by  —  20,  x  =  —  2. 

To  expand  an  algebraic  expression  is  to  perform  the  operations  indicated. 

From  the  above  examples,  we  have  the  following  rule  for 
solving  an  integral  linear  equation  with  one  unknown  number : 

Clear  the  equation  of  fractions,  if  any,  by  multiplying  each  term 
by  the  L.  C.  M.  of  the  denominators  of  the  fractional  coefficients. 

Remove  the  parentheses,  if  any,  by  performing  all  the  opera- 
tions indicated. 

Transpose  the  unknown  terms  to  the  first  member,  and  the 
known  to  the  second  ;  cancelling  any  term  which  has  the  same 
coefficient  in  both  members. 

Unite  similar  terms,  and  divide  both  members  by  the  coefficient 
of  the  unknown  number. 

The  pupil  should  verify  every  result. 

EXERCISE  23 

Solve  the  following  equations,  in  each  case  verifying  the 
answer : 

1.  8»  +  7  =  95.  8.  6x-2S  =  15x-lS. 

2.  9x  =  5x-S2,  9.  19-13a;  =  31-29a;. 

3.  7a;  +  15  =  2a;  +  45.  10.  14aj-51  =  27a;-33. 
>4.   10a;-3  =  3a;-38.  11.  13 +12  a;  =  37  a; +  43. 

5.  6a;  +  13  =  lla;-7.  V12.   21a;- 23  =  51 -16a;. 

6.  5-18a;  =  83-12a;.  13.   lla;  +  17  =  65a;  +  47. 
V7.  lla;-3  =  4  +  3a;.  14.  98-16a;  =  23-41a;. 


56 


ALGEBRA 


15. 
16. 
17. 

V18.   2x- 


17a;-9  4-47  =  41a;-35a;  +  27. 
13  a;- 39  =  48  a; -29  a; -81. 
54  =  26  a;-31a;  +  19a;-9. 


?a;  +  ia;  =  10. 
3        7 


21. 

22. 

7'"  +  4-n''  +  28'' 

23. 

2^     38     8^     4^ 

5^-i5=9*-r 

55 

o>i  5        7        7         1 

^-  6"-8"  =  9"-i8"-48 

i/25.  4(2a;-7)  +  5  =  5(a;-3)  +  16. 

V26.  13x-(5a;-8)  =  6a;-(3a;  +  7). 

27.  80-6(4x  +  3)  =  7a;-3(6a;  +  l). 

28.  x-2(4-7a;)  =  4a;-9(2-3a)). 

29.  8a;(3a;  +  2)-27  =  4a;(6a;-l)-147. 

30.  45-5a;(6aj-l)=21-3a;(10a;  +  3). 

31.  4(a; +  14) -4  (3 a; -32)  =  6(a;  + 12) -7(a;- 12). 
./32.  (5-3  a;)(3  +  4aj)  =  (7  +  3a;)(l  -4  a;)  - 1. 

V^33.  (l  +  3a;)2=(5-a;)2  +  4(l-a;)(3-2a;). 

34.  6(aj  +  4)2  =  5-(2a;  +  3)2-5(2-a;)(7  +  2a;). 

V35.  (3aj-2)2-9(a;-l)(3a;-8)  =  -18a^f 51a;-38. 

36.  (aj  +  4)3-(x-4)3  =  2(3a;-2)(4a.'  +  l). 
1 


|(4x  +  lJ  +  .f6. 


V: 


__-J^«9. 


)-l(5.  +  8) 
|(.-3)-|(...z)-X(._5)  = 


2. 


16 


2-3 


-) 


INTEGRAL   LINEAR  EQUATIONS  57 

PROBLEMS    LEADING   TO    INTEGRAL    LINEAR    EQUATIONS 
WITH  ONE  UNKNOWN  NUMBER 

91.  For  the  solution  of  a  problem  by  algebraic  methods,  the 
following  suggestions  will  be  found  of  service : 

1.  Represent  the  unknown  number,  or  one  of  the  unknown 
numbers  if  there  are  several,  by  some  letter,  as  x. 

2.  Every  problem  contains,  explicitly  or  implicitly,  just  as 
many  distinct  statements  as  there  are  unknowri  numbers  involved. 

Use  all  but  one  of  these  to  express  the  other  unknown  num- 
bers in  terms  of  x. 

3.  Use  the  remaining  statement  to  form  an  equation. 

92.  Problems. 

1.  Divide  45  into  two  parts  such  that  the  less  part  shall  be 
one-fourth  the  greater. 

Here  there  are  two  unknown  numbers  ;  the  greater  part  and  the  less. 
In  accordance  with  the  first  suggestion  of  §  91,  we  represent  the  greater 
part  by  x. 

The  first  statement  of  the  problem  is,  implicitly : 
The  sum  of  the  greater  part  and  the  less  is  45. 
The  second  statement  is  : 
The  less  part  is  one-fourth  the  greater. 

In  accordance  with  the  second  suggestion  of  §  91,  we  use  ih.Q  first  state- 
ment to  express  the  less  part  in  terms  of  x. 

Thus,  the  less  part  is  represented  by  45  —  x. 

We  now,  in  accordance  with  the  third  suggestion,  use  the  second  state- 
ment to  form  an  equation. 


Thus, 

45-a;  =  -x. 

Clearing  of  fractions, 

180  -  4  a;  =  x. 

Transposing, 

-  4  a;  -  X  =  -  180,  or  -  5  a;  =  -  180. 

Dividing  by  -  6, 

x  =  36,  the  greater  part. 

Then, 

45  —  X  =  9,  the  less  part. 

2.   A  had  twice  as  much  money  as  B;  but  after  giving  B 
^28,  he  has  f  as  much  as  B.     How  much  had  each  at  first? 


58  ALGEBRA 

Let  X  represent  the  number  of  dollars  B  had  at  first. 
Then,  2  x  will  represent  the  number  A  had  at  first. 
Now  after  giving  B  $28,  A  has  2  a;  -  28  dollars,  and  B  re  +  28  dollars ; 
we  then  have  the  equation 

2a;-28  =  ^(x  +  28). 

o 

Clearing  of  fractions,  6  a;  -  84  =  2  (a;  +  28) . 

Expanding,  6  x  —  84  =  2  a;  +  56. 
Transposing,  4  a:  =  140. 

Dividing  by  4,  x  =  35,  the  numberof  dollars B  had  at  first ; 

and  2  X  =  70,  the  number  of  dollars  A  had  at  first. 

3.  A  is  3  times  as  old  as  B,  and  8  years  ago  he  was  7  times 
as  old  as  B.     Eequired  their  ages  at  present. 

Let  X  =  the  number  of  years  in  B's  age. 

Then,  Sx=  the  number  of  years  in  A's  age. 

Also,  x  —  S=  the  number  of  years  in  B's  age  8  years  ago, 

and  3ft  —  8  =  the  number  of  years  in  A's  age  8  years  ago. 

But  A's  age  8  years  ago  was  7  times  B's  age  8  years  ago. 
Whence,  3a;_8  =  7(x-8). 

Expanding,  3x-8  =  7x-56. 

Transposing,  _  4  x  =  —  48. 

Dividing  by  —  4,  x  =  12,  the  number  of  years  in  B's  age. 

Whence,  3  x  =  36,  the  number  of  years  in  A's  age. 

4.  A  sum  of  money  amounting  to  $4.32  consists  of  108 
coins,  all  dimes  and  cents ;  how  many  are  there  of  each  kind  ? 

Let  X  =  the  number  of  dimes. 

Then,  108  —  x  =  the  number  of  cents. 

Also,  the  X  dimes  are  worth  10  x  cents. 
But  the  entire  sum  amounts  to  432  cents. 
Whence,  10  x  +  108  -  x  =  432. 

Transposing,  9  x  =  324. 

Whence,  x  =  36,  the  number  of  dimes  ; 

and  108  -  X  =  72,  the  number  of  cents. 


INTEGRAL  LINEAR  EQUATIONS  59 

EXERCISE  24 

sll.    The  difference  of  two  numbers  is  12,  and  7  times  the 
smaller  exceeds  the  greater  by  30.     Find  the  numbers. 

"^  2.  The  sum  of  two  numbers  is  29,  and  the  smaller  exceeds 
their  difference  by  4.     Find  the  numbers. 

3.  Find  two  numbers  whose  sum  is  -J,  and  difference  ^. 

4.  The  sum  of  two  numbers  is  44,  and  their  difference  is 
three-fourths  the  smaller  number.     Find  the  numbers. 

^5.  A  is  4  times  as  old  as  B ;  and  in  22  years  he  will  be  twice 
as  old. '  Find  their  ages. 

6.   A  is  3  times  as  old  as  B ;  and  6J  years  ago  he  was  5  times 
as  old.     Find  their  ages. 

J  7.  A  has  3  times  as  much  money  as  B ;  but  after  B  gives 
him  $9,  he  has  6  times  as  much  as  B.  How  much  had  each 
at  first? 

^  8.  A  man  has  21  ,  coins,  all  dimes  and  twenty-five-cent 
pieces,  valued  in  all  at  $3.30.     How  many  has  he  of  each? 

9.   A  is  25  years  of  age,  and  B  is  16.     In  how  many  years 
will  B  be  t\^o-thirds  as  old  as  A  ? 

\i  10.  Divide  43  into  two  parts  such  that  if  the  greater  be 
added  to  17,  and  the  less  to  30,  the  resulting  numbers  shall 
be  equal. 


11.  Twice  a  certain  number  exceeds  35  by  the  same  amount 
that  one-third  the  number  exceeds  5.     Find  the  number. 

12.  Divide  $280  between  A,  B,  and  C  so  that  A's  share  may 
exceed  f  of  B's  by  $96,  and  B's  share  exceed  C's  by  $20. 

13.  A  is  22  years  of  age,  and  B  is  18.  How  many  years  ago 
was  A's  age  f  of  B's  ? 

14.  A  man  has  $4.10,  all  five-cent  and  fifty-cent  pieces; 
and  he  has  5  more  five-cent  than  fifty-cent  pieces.  How  many 
has  he  of  each  ? 


60  ALGEBRA 

15.  The  sum  of  |  and  f  a  certain  number  exceeds  |-  the 
number  by  |.     Find  the  number. 

16.  If  A  has  $  5.50,  and  B  ^3.50,  how  much  money  must  A 
give  B  in  order  that  B  may  have  |  as  much  as  A  ? 

V  17.  A  room  is  f  as  long  as  it  is  wide;  if  the  width  were 
increased  by  1^  feet,  and  the  length  diminished  by  the  same 
amount,  the  room  would  be  square.     Find  its  dimensions. 

18.  The  sum  of  two  numbers  is  {^  the  greater,  and  their  dif- 
ference is  I.     Find  the  numbers. 

V 19.  A  boy  buys  a  certain  number  of  apples  at  2  for  5  cents, 
and  double  the  number  at  3  for  5  cents,  and  spent  in  all  35 
cents. .   How  many  of  each  kind  did  he  buy  ? 

20.  Divide  $320  between  A,  B,  C,  and  D  so  that  A  may 
receive  $  35  more  than  B,  C  $  15  more  than  B,  and  D  $  25  less 
than  C. 

21.  The  sum  of  the  ages  of  A,  B,  and  C  is  52  years ;  A's  age 
is  f  of  B's,  and  he  is  8  years  younger  than  C.    Find  their  ages. 

22.  In  a  certain  school  the  boys  are  15  fewer  than  |  of  the 
whole,  and  the  girls  are  33  more  than  ^.  How  many  boys,  and 
how  many  girls,  are  there  ? 

I  23.  The  sum  of  $900  is  invested,  part  at  4%,  and  the  rest 
at  5%,  per  annum,  and  the  total  annual  income  is  $42.  How 
much  is  invested  in  each  way  ? 

1^24.  In  9  years  B  will  be  f  as  old  as  A;  and  12  years  ago  he 
was  I  as  old.     What  are  their  ages  ? 

Let  X  represent  the  number  of  years  in  A's  age  12  years  ago. 

25.  A  has  I  of  a  certain  sum  of  money,  B  has  ^,  C  has  |, 
and  D  has  the  remainder,  $8.     How  much  have  A,  B,  and  C  ? 

V  26.  A  man  bought  8  hens,  7  sheep,  and  12  pigs  for  $269; 
each  sheep  cost  -y-  as  much  as  each  hen,  and  $  3  less  than  each 
pig.     What  did  each  cost  ? 

27.  Divide  66  into  two  parts  such  that  f  the  greater  shall 
exceed  f  the  less  by  21. 


INTEGRAL  LINEAR  EQUATIONS  61 

i/  28.  Find  two  numbers  whose  sum  is  10,  such  that  the  square 
of  the  greater  exceeds  the  square  of  the  less  by  40. 

29.  Find  two  consecutive  numbers  such  that  ^  the  greater 
exceeds  I  the  less  by  2. 

1^30.  A  person  attempting  to  arrange  a  certain  number  of 
counters  in  a  square  finds  that  he  has  too  few  by  12;  but  on 
reducing  the  number  in  the  side  of  the  square  by  3,  he  has 
21  left  over.     How  many  has  he  ? 

31.  A  purse  contains  a  certain  number  of  10-shilling  pieces, 
twice  as  many  5-shilling  pieces,  and  5  times  as  many  shillings, 
the  contents  of  the  purse  being  worth  £5.  How  many  are 
there  of  each  coin  ? 

32.  The  square  of  the  third  of  three  consecutive  numbers 
exceeds  the  product  of  the  other  two  by  13.   Find  the  numbers. 

33.  Divide  39  into  two  parts  such  that  3  times  the  smaller 
shall  be  as  much  below  58  as  twice  the  greater  exceeds  38. 

34.  Find  two  numbers  whose  difference  is  3,  and  whose 
product  is  less  by  33  than  the  square  of  the  greater. 

''  35.  The  total tiumber  of  persons  at  a  certain  factory  is  196; 
the  number  of  women  is  f  the  number  of  men,  and  |  the  num- 
ber of  boys.     How  many  of  each  are  there  ? 

36.  A  room  is  twice  as  long  as  it  is  wide,  and  it  is  found 
that  50  square  feet  of  carpet,  1  foot  in  width,  are  required  to 
make  a  border  around  it.     Find  its  dimensions. 

37.  A  purse  contains  a  certain  number  of  dimes,  |-  as  many 
cents,  and  ^  as  many  $  1  bills,  the  value  of  the  entire  contents 
being  $  5.74.     How  many  are  there  of  each  ? 

38.  A  starts  to  walk  from  P  to  Q,  12  miles,  at  the  same 
time  that  B  starts  to  walk  from  Q  to  P.  They  meet  at  the 
end  of  2  hours.  If  A  walks  one  mile  an  hour  faster  than  B, 
what  are  their  rates  ? 

39.  Divide  ^210  between  A,  B,  C,  and  D  so  that  B  may  re- 
ceive $  10  less  than  A,  G  y-  as  much  as  B,  and  D  -|  as  much  as  A. 


62  ALGEBRA 

40.  The  sum  of  $  32  is  divided  between  7  men,  8  women, 
and  16  children ;  each  child  receiving  i  as  much  as  each  man, 
and  each  woman  75  cents  more  than  each  child.  How  much  is 
received  by  each  man,  each  woman,  and  each  child  ? 

41.  A  boy  had  a  certain  number  of  marbles.  He  lost  6  of 
them,  gave  away  ^  the  remainder,  and  then  found  that  he 
had  5  more  than  i  of  his  original  number.  How  many  had 
he  at  first  ? 

42.  There  are  two  heaps  of  coins,  one  containing  5-cent 
pieces  and  the  other  10-cent  pieces.  The  second  heap  is  worth 
20  cents  more  than  the  first,  and  has  8  fewer  coins.  Find  the 
number  in  each  heap. 

43.  In  an  audience  of  435  persons,  there  are  25  more  women 
than  men,  and  3  times  as  many  girls  as  men ;  and  the  number 
of  boys  is  less  by  195  than  twice  the  number  of  girls.  Find 
the  number  of  each. 

{.  44.  Find  four  consecutive  odd  numbers  such  that  the  prod- 
uct of  the  first  and  third  shall  be  less  than  the  product  of  the 
second  and  fourth  by  64. 

,"  45.  A  sum  of  money,  amounting  to  $19.30,  consists  of  $2 
bills,  25-cent  pieces,  and  5-cent  pieces.  There  are  13  more 
5-cent  pieces  than  $  2  bills,  and  |-  as  many  5-cent  pieces  as 
25-cent  pieces.     Find  the  number  of  each. 

46.  Two  barrels  contain  46  aijd  45  gallons  of  water,  respec- 
tively. A  certain  number  of  gallons  are  drawn  from  the  first, 
and  I  as  many  from  the  second,  and  the  second  now  contains 
f  as  many  gallons  as  the  first.  How  many  gallons  were  drawn 
from  each  ?        . 

47.  A  tank  containing  150  gallons  can  be  filled  by  one  pipe 
in  15  minutes,  and  emptied  by  another  in  25  minutes.  After 
the  first  pipe  has  been  open  a  certain  number  of  minutes,  it  is 
closed,  and  the  second  pipe  opened;  and  the  tank  is  emptied 
in  24  minutes  from  the  time  the  first  pipe  was  opened.  How 
many  minutes  is  each  pipe  open  ? 


SPECIAL  METHODS  63 


VII.     SPECIAL    METHODS    IN    MULTIPLICA- 
TION AND  DIVISION 

93.  Any  Power  of  a  Power. 

Required  the  value  of  {a^y. 

By  §  11,  {ay  =  a'xa'xa^  =  a\ 

We  will  now  consider  the  general  case : 

Eequired  the  value  of  (ccy,  where  m  and  n  are  any  positive 
integers. 

We  have,        (a™)"  =  ar  xoT  X  -"  ton  factors 

^m+m+"-  to  n  terms  __  Qj^n 

94.  Any  Power  of  a  Product. 

Eequired  the  value  of  (abf. 

By  §  11,  (ahf  =  abxabxab  =  a%K 

We  will  now  consider  the  general  case : 
«    Eequired  the  value  of  (a&)",  where  n  is  any  positive  integer. 

We  haye,     (aby  =ab  xab  x-"  to  n  factors  =  a''b\ 

In  like  manner,       (abc  •.•)"  =  a^'b^'c^  •  •  •, 
whatever  the  number  of  factors  in  abC": 

95.  Any  Power  of  a  Monomial. 

1.  Find  the  value  of  (-5a*y. 

By  §31,  (_5a4)3=[(-5)xa4]3 

=  (-  5)8  X  («4)8(§  94)  =  -  125ai2(§93). 

2.  Find  the  value  of  (-  2  u^ny. 

,  We  have,     (  -  2  m^ny  =  (  -  2)*  x  (m^)*  x  w*  =  16  wi^w*. 


64  ALGEBRA 

.  96.  From  §§93  and  94  and  the  examples  of  §  95,  we  have 
the  following  rule  for  raising  a  rational  and  integral  monomial 
(§  63)  to  any  power  whose  exponent  is  a  positive  integer : 

Raise  the  absolute  value  of  the  numerical  coefficient  to  the 
required  power,  and  multiply  the  exponent  of  each  letter  by  the 
exponent  of  the  required  power. 

Give  to  every  power  of  a  positive  term,  and  to  every  even  power 
of  a  7iegative  term,  the  positive  sign ;  and  to  every  odd  power  of 
a  7iegative  term  the  negative  sign. 

EXERCISE  25 

Expand  the  following : 

1.  (a^y*z^y.                5.    (Ja^b^^'f.  9.  (a^b^cy. 

2.  (m^n'py\               6.    (- nV/)i«.  10.  (a^^y^z^y\ 

3.  (-ab'cy.            7.    (2m'xy.  11.  (-Sm^n^x^. 

4.  (^  11  a^fy.          8.    (-4a^2/0'-  12.  (-2amV)^ 

97.   Square  of  a  Binomial. 

Let  it  be  required  to  square  a-\-b. 

a  +  b 


a^  +  ab 

ab-\-b' 
Whence,  (a  -\-by  ==  a^ +  2  ab  +  b\  (1) 

That  is,  the  square  of  the  sum  of  two  numbers  equals  the  square 
of  the  first,  plus  twice  the  product  of  the  first  by  the  second,  plus 
the  square  of  the  second, 

1.    Square  3  a +  2  &. 

We  have,      (3  a  +  2  6)2  =  (3  aY  +  2(3  a)  (2  6)  +  (2  6)2 

=  9a2  +  i2a6  +  4  62. 

Let  it  be  required  to  square  a  —  b. 


SPECIAL  METHODS  65 

a  —  b 
a  —  b 


a^    —ab 

-  a6  +  b\ 
Whence,  {a-bf  =  a' -2  ab  +  b\  (2) 

That  is,  the  square  of  the  difference  of  tivo  numbers  equals  the 
square  of  the  first,  minus  twice  the  product  of  the  first  by  the  sec- 
ond, plus  the  square  of  the  second. 

In  the  remainder  of  the  work  we  shall  use  the  expression  "  the  differ- 
ence of  a  and  &"  to  denote  the  remainder  obtained  by  subtracting  h  from  a. 

The  result  (2)  may  also  be  derived  by  substituting  —  6  for  6,  in  equa- 
tion (1). 

2.  Square  4  a?^  —  5. 

We  have,  (4  x^  -  5)2  =  (4  ^2)2  _  2(4  x"^)  (5)  +  52 

=  16  x*  -  40  a;2  +  25. 

If  the  first  term  of  the  binomial  is  negative,  it  should  be  en- 
closed, negative  sign  and  all,  in  parentheses,  before  applying 
the  rules. 

3.  Square  -2a^  +  9.  ^  "^  ^^ 
We  have,       (_  2  a^ -f  9)2  =  [(- 2  a^)  +  gp 

'    =  (-2a3)2  +  2(-2a3)(9)4.92 
=  4  a6  _  36  a3  +  81.   . 

EXERCISE   26 

Expand  the  following : 

1.  {a^-2)\  7.  (7a;-f3a^l  13.  (4  a^ - 11 .7^)2. 

2.  (a; -5)2.  8.  (^n^  +  llny.  14.  {^  ax -12  byf. 

3.  {Qx-^lyf.  9.  (2a«-7  62c)2.  15.  {-^n'  +  lOn^. 

4.  (3  +  8n2)2.  10.  (_4m4-3n3)2.  16.  (8a;^  +  9a;y. 

5.  (_m^  +  4p^y.     11.  {Qofy  +  x'yy.  17.  (7  a^m^ - 13  6 V) I 
/6.    (9a6-l)l         12.  (5a6  4-8  6c)2.  18.  (-6a^-ll  a;2;)2. 

19.    {bx^  +  ^y^y.  20.  (2a--9a'')l 


66  ALGEBRA 

98.   Product  of  the  Sum  and  Difference  of  Two  Numbers. 

Let  it  be  required  to  multiply  a  +  6  by  a  —  6. 


o?  +  ah 
-ab-b' 
Whence,  (a -{-b)(a-b)  =  a'  -b\  . 

That  is,  the  product  of  the  sum  and  difference  of  two  numbers 
equals  the  difference  of  their  squares. 

1.  Multiply  6a4-563  by  6a-56l 
By  the  rule, 

(6  a  +  5  63)  (6  a  -  5  63)  =  (6  ay  -  (5  63)2  =  36  a2  -  25  5», 

2.  Multiply  -a^  +  4by-a;2_4 

(-  x2  +  4)  (-  x2  -  4)  =  [(-  x2)  +  4]  [(-  a:2)  -  4] 
=:(_a;2)2_42  =  aj4-16. 

3.  Expand  (x  +  y  +  z)(x  —  y-{-z). 

ix  +  y  +  z)ix-y  +  z)  =  [(x  +  z)  +  y;\  [(x  +  z)-y^ 

=  (X  +  0)2  -  1/2 

=:X^  +  2XZ  +  Z^-  2/2. 

4.  Expand  (a  +  5  —  c)  (ct  —  &  +  c). 

By  §  52,   (a  +  6  -  c)  (a  -  6  +  c)  =  [a  +  (6  -  c)]  [a  -  (6  -  c)] 

=  a2  -  (6  -  c)2,  by  the  rule, 
=  a2  _  (52  _  2  6c  +  c2) 
=  ^2  -  62  +  2  6c  -  c2. 

EXERCISE  27 

Expand  the  following : 

1.  (7a  +  2b)(7a-2b).  V3.   (3a;3  +  8  2/;22)(3a53_g2^^2)^ 

2.  (9m2  +  4)(9m2-4).  4.   (-a«  +  6)(-a3-6). 


SPECIAL  METHODS  67 

5.   (llm'-{-5n')(llm'-5n').   7.  (5  a' +  12  b^iBa' -12 b^. 

■'  9.   (-10  m*?i+ 13  a^) (- 10  m%  -  13  a^). 

10.  (a-b-{-c){a-b-c).  13.  (1 +a- &)(1 -a  +  &). 

11.  {x'-{-x  +  l){x'  +  x-l).      '14.  (a2  +  3a+l)(a2-3a  +  l). 

12.  (a;  +  2/  +  ^)(«-2/-2;)-  15.  (x  +  y +  ^){x-y -'d). 

16.  (a;2  +  a^  +  2/2)(a;2-a;2/  +  2/^). 

17.  .(a'  +  5a-4)(a2-5a  +  4). 

18.  (4ic2  +  3a;-7)(4x2-3aj-7). 

V  19.  (m^  +  5  m^n^  +  2  n^)  (m^  -  5  mV  -  2  ti^). 

99.   Product  of  Two  Binomials  having  the  Same  First  Term. 

Let  it  be  required  to  multiply  x-\-a'bj  x-\-b. 

x-{-a 
x  +  b 


3?-\-  ax 

+  bx-\-ah 

Whence,       (x  +  a)  (a;  +  6)  =  x^  H-  (a  +  b)x  +  ab. 

That  is,  the  product  of  two  binomials  having  the  same  first  term 
equals  the  square  of  the  first  ternV)  plus  the  algebraic  sum  of  the 
second  terms  multiplied  by  the  first  term,  plus  the  product  of  the 
second  terms. 

1.  Multiply  a;  —  5  by  a;  +  3. 

The  coeflB.cieut  of  x  is  the  sum  of  —  5  and  +  3,  or  —  2. 
The  last  term  is  the  product  of  —  5  and  +  3,  or  —  15. 

Whence,  (x  -  5)(x  +  S)  =  x^  -  2x  -  15. 

2.  Multiply  a; -5  by  a; -3. 

The  coefficient  of  x  is  the  sum  of  —  5  and  —  3,  or  —  8. 
The  last  term  is  the  product  of  —  5  and  —  3,  or  15. 

Whence,  («  -  5)(x  -  3)  =  x^  -  8  x  +  15. 


68  ALGEBRA 

3.  Multiply  ab  —  4:  by  ab-{-7. 

The  coefficient  of  ab  is  the  sum  of  —  4  and  7,  or  3. 
The  last  term  is  the  product  of  —  4  and  7,  or  —  28. 

Whence,  (ab  -  4)  (a&  +  7)  =  ^252  ^Sab-  28. 

4.  Multiply  x^-\-6y^  hy  x^-\-Sy\ 

The  coefficient  of  x^  is  the  sum  of  6  y^  and  8  y^,  or  14  y^. 
The  last  term  is  the  product  of  6  y^  and  8  y^,  or  48  y^. 

Whence,      (x^  +  6  y^)  {x^  -\-8y^)  =  x*-^  14  x^^  +  48  ^. 

EXERCISE  28 
Expand  the  following  by  inspection : 
1.    (x  +  S)(x  +  4:).  m.    (x-y  +  7)(x-y-e), 

12.    (x-2)(x-\-5).  12.    (a  +  Sx)(a  +  9x). 

3.   (x-ll)ix-l).  13.   (x-9y)(x-5y). 

^4.   (a-7)(a  +  2).  v  14.   (m^-f-GyiXm^-Tw). 

V  5.    (a2  +  15)(a'  + 1).  V  15.    (a  +  6  +  2)(a  +  &  +  13). 

6.    (m3-3)(m3  +  8).  16.    (x"--{-10y^^)(x'"' -9f-). 

J.    (x--2)(x--6).  17.   (a^-9b%a'  +  Sb*). 

8.  (a'"  +  10)(a'"  +  2).  18.    (m/i- 14a;?/)(m7i-4a;2/). 

9.  (mn-7)(mn-3).  V19.    (7/1-71- 3)(m -n +  11). 
VlO.    (ab  +  l){ab-3).  20.    (a^ft +  11  c3)(a26- 12 c^). 

ICX).  Product  of  Two  Binomials  of  the  Form  mx  +  n  and  px  +  g. 
We  find  by  multiplication : 


mx-{-n 

X 

px-\-q 


mpa?  +  npic 

+  mqx-\-nq 

mpx^  +  (?ip + mq)  x-\-nq. 


SPECIAL  METHODS  69 

The  first  term  of  this  result,  mpx^,  is  the  product  of  the  first 
terms  of  the  binomial  factors,  and  the  last  term,  nq,  the  product 
of  the  second  terms. 

The  middle  term,  {np  +  mq)x,  is  the  sum  of  the  products  of 
the  terms,  in  the  binomial  factors,  connected  by  cross  lines. 

Ex.     Multiply  3aj  +  4by2aj  —  5. 

The  first  term  is  the  product  of  3  x  and  2  ic,  or  6  aj2. 
The  coefficient  of  x  is  the  sum  of  4  x  2  and  3  x  (—  5)  ;  that  is,  8  —  15, 
or  —  7. 

The  last  term  is  the  product  of  4  and  —  5,  or  —  20. 

Whence,  (35c  +  4)(2a;  -  5)  =  6x2  -  7ic  -  20. 

EXERCISE  29 
Expand  the  following  by  inspection : 

1.  (x  +  6)(3x  +  2).  9.    (2aa;-3)(5aa;  +  6). 

2.  (2x^l)(lx-l).  10.    (3a?4-2n)(10a;-n). 

3.  (2x-5)(4aj  +  3).  11.   (4aj-32/)(9a;  +  2y). 

4.  (4a-3)(5a-3),  12.    (7a- 2m)(7a-4m). 

5.  (4m  +  l)(4m  +  3).  13.    {^x"" -\-y){^x'' ^y). 

6.  (3  7i  +  2)(57i-2).  14.    (6a2  +  aj2)(8a2-5a^). 

7.  (2a2-l)(lla2-4)  16.    {^m^ -2n%10'm?-7  n"), 

8.  (5aj^  +  6)(6a;4+l).  16.    {^ax-Zhy){^ax  +  bhy)'. 

17.  [6(m  +  w)-5][(m  +  n)-2]. 

18.  [3(a-&)-|-4][4(a-6)-3]. 

101.  We  find  by  division, 

-— —  =  a-6-  --  =  a4-6, 

a-\-o  a  —  o 

That  is. 

If  the  difference  of  the  squares  of  two  numbers  be  divided  by  the 
sum  of  the  numbers,  the  quotient  is  the  difference  of  the  numbers. 


70  ALGEBRA 

If  the  difference  of  the  squares  oftivo  numbers  be  divided  by  the 
difference  of  the  numbers,  the  quotient  is  the  sum  of  the  numbers. 

1.  Divide  25  yh'  -9hj5yz'-  3. 

By  §  96,  25  y^si^  is  the  square  of  5  yz^  ;  then,  by  the  second  rule, 

^^y'''-^  =  5yz^  +  S. 
byz^-'6  ■ 

2.  Divide  x^  —  {y—zf  by  x-{-{y  —  z). 

By  the  first  rule,  ^^  " Cy  -  ^^  =  x -(y  -  z)  =  x  -  y  +  z. 

x  +  iy  -z) 

EXERCISE  30 

Find,  without  actual  division,  the  values  of  the  following: 
S6a'^-121b*'^         n    225  a^'- 100  b^^ 


VI.   "^ — ^.  vSi. 


a2-4 

a  +  2 

a^-9 

x-S 

25  n^ - 

-1 

5n'- 

1 

16  x^ - 

■81 

6aP-llb^^  '      15a«  +  10 


8n^H-ar'  '  *       14mn^-16a;« 

v^*3    25n^-l  y    l-144a^^&^         ^^    4a^-(6-cV 

'     "    "      '  '  '1-12  a"6^  *  .  *    2  a  -  (6  -  c)  * 

.     16a;^-81         g    49.rV°-1692^      ,«    a^-(m  +  3n)^ 
'      4a^  +  9   *    ^'    '      7a^2/^  +  132*    '         *    a  +  (m  +  3w)* 

V^  13    (q  +  a;y-(&-y)' 
(a  +  a;)  +  (&-2/) 

102.  We  find  by  division, 

a+b  a—b 

That  is, 

ijf  the  sum  of  the  cubes  of  two  numbers  be  divided  by  the  sum 
of  the  numbers,  the  quotient  is  the  square  of  the  first  number, 
minus  the  product  of  the  first  by  the  second^  plus  the  square 
of  the  second  number. 


SPECIAL  METHODS  71 

If  the  difference  of  the  cubes  of  two  numbers  be  divided  by  the 
difference  of  the  numbers,  the  quotient  is  the  square  of  the  first 
number,  plus  the  product  of  the  first  by  the  second,  plus  the  square 
of  the  second  number. 

1.  Divide  l  +  Sa^  by  l4-2a. 

By  §  96,  8  a^  is  the  cube  of  2  a  ;  then,  by  the  first' rule, 

l+2a         l+2a  ^     ^ 

2.  Divide  27 x'-64.y^  by  3x'-4:f. 
By  the  second  rule, 

27xe-64y9^(3.^)3-(4yB)8^ 
3x2-4^/3  3x2-4?/3  ^       ^   ^^       ^^   ^ ''^^   ^  ' 

=  9  x*  +  12  xV  +  16  y«. 

EXERCISE  31 

Eind,  without  actual  division,  the  values  of  the  following : 

•  1    ^  +  1  g    £^!±A'.  11     27a;«-125i/^ 

*  ic  +  l*  '   a=^  +  62*  •       Zx^-hy 

2    ^-^'  ^  7    ^^±125.  12    343  m^r^^  + 8 1^3 

*  1  — a  *      a +  5  *       7mn4-2p 

•  3    ^'-27  g    64  a?^"*-!  ^3    64a^6^  +  216c^ 

*  n-3*  *    4a^--J  *  *       4a26  +  6c3 

^  4    8  +  m^  ,^  g    a^6«-216  ^14    a;^«  - 1000  y^g^^ 

24-m^*  *      a6-6    *  *       a;«-102/V 

5    a;V^-g9  ,Q    343m«  +  n«  ^^5    729ftV4-512  3/« 

*  xY-^'  '      lm?  +  n  *  *        9a3aj  +  82/' 

103.  We  find  by  actual  division, 
a'-b' 


a-\-b 


a^-a'b-\-ab'-l^. 


^  =  a'  +  a'b-^ab^-\-b^ 


72  •        ALGEBRA 

a+b 
^'-^'     a^^a'b-{-a'b'  +  a¥-\-b'',  etc. 


a  —  b 
In  these  results,  we  observe  the  following  laws : 

I.  The  exponent  of  a  in  the  Jlrst  term  of  the  quotient  is  less  by 
1  than  its  exponent  in  the  dividend,  and  decreases  by  1  in  each 
succeeding  term. 

II.  Tlie  exponent  of  b  in  the  sec&nd  term  of  the  quotient  is  1, 
and  increases  by  1  m  each  succeeding  term. 

III.  If  the  divisor  is  a—b,  all  the  terms  of  the  quotient  are 
positive;  if  the  divisor  is  a-\-b,  the  terms  of  the  quotient  are 
alternately  positive  and  negative. 

A  general  proof  of  these  laws  will  be  found  in  §  466. 

1.  Divide  a^  —  b''  hj  a  —  b. 
By  the  above  laws, 

a  —  b 

2.  Divide  16  a;^- 81  by  2  a; +  3. 

We  have,         16  o^^  -  81  ^  ^2  .)^  -  3^ 

2a;  +  3  2  a;  +  3 

=  (2  x)3  -  (2  a;)2  .  3  +  2  X  .  32  -  33 
=  8  a;3  _  12  a:2  +  18  X  -  27. 


EXERCISE  32 

Find,  without  actual  division,  the  values  of  the  following : 

1     ^*  ~  ^*  3    ^^~  ^.  5    a^  — /, 

a—b  '    x  —  1  '    (xf  +  y^' 

m  +  w'  ^    1  +  a'  ^^^    a«-6c2* 


SPECIAL  METHODS 

73 

"■  ?5f- 

15. 

m^-729n^ 

J^-   7/1^" -81 

©fi^fi*- 

i^ 

3a^-2/ 

9     «^-^'. 
a  — ic 

j3     256a5^-2/r 

17. 

a>  +  128  6^* 
a  +  262 

10.  »'  +  !. 

71  +  1 

..     a'  +  2^Sx'^ 
'       a  +  3a^- 

18. 

64a;«-729 
2a;-3 

104.   The  following  statements  will  be  found  to  be  true  if  n 

is  any  positive  integer : 

I.  a"  —  b""  is  always  'divisible  by  a  —  b. 

Thus,  a^  —  b^,  a^  —  b%  a*  —  b%  etc.,  are  divisible  hy  a  —  b. 

II.  a^  —  b""  is  divisible  by  a-\-b  ifn  is  even. 

Thus,  a-  —  b^,  a'*  —  b^,  a^  —  b^,  etc.,  are  divisible  by  a  4-  6. 

III.  a**  +  6"-  /s  divisible  by  a-\-b  ifn  is  odd. 

Thus,  a^  +  b^,  a^  +  &'^,  a'  +  6'',  etc.,  are  divisible  by  a  +  6. 

IV.  a"*  +  6**  IS  divisible  by  neither  a-^b  nor  a  —  b  if  n  is 


even. 


Thus,  a^-{-b%  a^  +  6^,  a^  +  b^,  etc.,  are  divisible  by  neither 
a  +  b  nor  a  —  b. 

Proofs  of  the  above  statements  will  be  found  in  §  467. 


74  ALGEBRA 


VIII.    FACTORING 

105.  To  Factor  an  algebraic  expression  is  to  find  two  or  more 
expressions  which,  when  multiplied  together,  shall  produce  the 
given  expression. 

In  the  present  chapter  we  consider  only  the  separation  of  rational  and 
integral  expressions  (§  63),  with  integral  numerical  coefficients,  into  fac- 
tors of  the  same  form. 

A  Common  Factor  of  two  or  more  expressions  is  an  expression 
which  will  exactly  divide  each  of  them. 

106.  It  is  not  always  possible  to  factor  an  expression ;  there 
are,  however,  certain  forms  which  can  always  be  factored; 
these  will  be  considered  in  the  present  treatise. 

107.  Case  I.  When  the  terms  of  the  expression  have  a  com- 
mon factor. 

1.  Factor  14  a6^- 35  a«62. 

Each  term  contains  the  monomial  factor  7  ah\ 
Dividing  the  expression  by  7  ah'^^  we  have  2  62  _  5  a\ 
Then,  14  a6*  -  35  a^&a  _  7  ^^i  (2  &2  _  5  «2). 

2.  Factor  (2  m +  3)^-2  + (2  m  4- 3)  2/'. 

The  terms  have  the  common  binomial  factor  2  m  +  3. 
Dividing  the  expression  by  2  m  +  3,  we  have  x'^  +  y\ 

Then,  (2  m  +  3)  x2  +  (2  m  +  3)  y3=  (2  w  +  3)  {pi?  +  y^). 

3.  Factor  (a  —  h)  m  -\-  (h  —  a)  n. 

By  §  52,  5  _  a  =  _  (a  -  &). 

Then,       (a  -  6)  m  +  (&  -  a)  w  =  (a  -  6)  wi  -  (a  -  6)  » 

=  (a-6)(TO-w). 
We  may  also  solve  Ex.  3  as  follows : 
(a  -  6)  TO  +  (6  -  a)  w  =  (6  -  a)  n  -  (6  -  a)  m  =  (J)'-a){n-'  m). 


FACTORING  75 

4.   Factor  5a(x  —  y)—Sa(x-{'y). 

=  a(6x  — 5y —  Sx  —  Sy) 
I  =a(2a:-8y)=2a(x-42/). 

EXERCISE  33 

Factor  the  following  : 
^  1.   63x^-54  3;^  5.   (a-2)6^- (a-2yd^ 

w  2.   a'-5a'-2a'  +  3a\  6.    (3  a;  +  5)m+ (3  a;  +  5). 

3.  m^n^  +  m^n"^  —  mn®.  i  7.    (m  —  n)  {x  +  y)  —  {n  —  m)  z. 

4.  24  a^2/' -  40  a;y  +  56  icY        8.   a(a2-2) +3(2-a2). 

^'9.    (x  4-  2/)  (m  4-  ^)  +  (a?  +  2/)  W  —  *^)- 
^10.   a(&  +  c)-a(5-c).  13.   5(2ic-i/J) -5(a;  +  32/). 

«*ll.   3aj2(a;-l)-i-(l-a;).  14.    (a  +  m)2-3(a  +  m). 

12.   6(3a  +  46)  +  6(5a-26).     15.   x2(52/-2;3) -a^(22/  +  ;2). 

16.  (m-n)34-2m(m-ny. 

17.  3a^+'-7a»+'b  +  a^. 

V  18.    (a-&)(m2  +  a;2)-(a-6)(m2-?/;2). 
19.    (m-ny~2m{m- nf  Jfrn^im- nf. 

108.  The  terms  of  a  polynomial  may  sometimes  be  so 
arranged  as  to  show  a  common  binomial  factor;  and  the  ex- 
pression can  then  be  factored  as  in  §  107. 

1.  Factor  ah  —  ay  +  hx  —  xy. 

By  §  107,      ah  -  ay  -\-hx~xy  =  a{'b-y)  +  x{h  -  y). 
The  terms  now  have  the  common  factor  b  —  y. 

Whence,       ab  —  ay  -{-bx —  xy  =  (a  +  x)(b  —  y). 

2.  Factor  a3  +  2a2-3a-6. 

If  the  third  term  is  negative,  it  is  convenient  to  enclose  the  last  two 
terms  in  parentheses  preceded  by  a  —  sign. 


76  ALGEBRA 

Thus,    a^-\-2a^-3a-6  =  (a^  +  2a'^)-(Sa  +  6) 

=  a2(a  +  2)  -  3(a  +  2)  =  (a2  -  3)(a  +  2). 

EXERCISE  34 

Factor  the  following : 
^  1.   ac -t  ad -\- be -\- bd.  V"3.   mx -\- my  —  nx —ny. 

3.  xy  —  Zx-{-2y  —  -&.  4.   a6  — a  — 56  +  5. 

5.  8  0^2/  + 12  ay  +  10  6a;  + 15  ab. 

6.  m*  + 6 m^—Tm  — 42. 

7.  6-10a  +  27a2-45a^ 

8.  20a6-28ad|-5  6c  +  7cd' 

9.  m^  —  m^7i  +  mn^  —  n^. 

10.  a^6^-a35Vd3-a25Vd2_|_c6^6^ 

11.  63  +  36x^  +  560^  +  320^. 

12.  48  a;?/ + 18  ria;  —  88  mi/ —  33  m?i. 

13.  mx  +  my  +  nx  +  ny  +px  +  py. 
,  14.  ax  —  ay  +  az  —  bx  +  by  —  bz. 

15."  3 am  —  6 an  +  Abm  —  S  b7i  +  cm  —  2 en. 
V 16.   ax  +.ay  —az—  bx  —  by  +  bz  +  cx  +  cy  —  cz. 

109.  If  an  expression  can  be  resolved  into  two  equal  fac- 
tors, it  is  said  to  be  a  perfect  square,  and  one  of  the  equal 
factors  is  called  its  square  root. 

Thus,  since  9  a'^b^  is  equal  to  3  a^6  X  3  a^b,  it  is  a  perfect 
square,  and  3a^6  is  its  square  root. 

9  05*62  is  also  equal  to  (—  3  a%)  x  (—  3  a^b);  so  that  —  3  a26  is  also  its 
square  root ;  in  the  examples  of  the  present  chapter,  we  shall  consider 
the  positive  square  root  only. 

110.  The  following  rule  for  extracting  the  positive  square 
root  of  a  monomial  perfect  square  is  evident  from  §  109 : 


FACTORING  77 

j  Extract  the  square  root  of  the  numerical  coefficient,  and  divide 
the  exponent  of  each  letter  by  2. 

Thus,  the  square  root  of  25  a'^6V  is  5  a^b^c. 

111.  It  follows  from  §  97  that  a  trinomial  is  a  perfect  square 
when  its  first  and  last  terms  are  perfect  squares  and  positive, 
and  the  second  term  plus  or  minus  twice  the  product  of  their 
square  roots. 

Thus,  in  the  expression  4:  x- — 12  xy  -\- 9  y^,  the  square  root  of 
the  first  term  is  2  x,  and  of  the  last  term  3  y ;  and  the  second 
term  is  equal  to  —  2  (2  x)  (3  y). 

Whence,  4:  x^  —  12  xy  -\- 9  y^  is  a,  perfect  square. 

112.  To  find  the  square  root  of  a  trinomial  perfect  square, 
we  reverse  the  rule  of  §  97 : 

j      Extract  the  square  roots  (§  110)  of  the  first  and  third  terms^ 
and  connect  the  results  by  the  sign  of  the  second  term. 

1.  Find  the  square  root  of  4:  x^  -{- 12  xy -^  9  y^. 
By  the  rule,  the  result  is2x-\-Sy. 

(The  expression  may  be  written  in  the  form 

(_2x)2  +  2(-2a:)(-3y)  +  (-3  2/)2, 

which  shows  that  (— 2  ic)  +  (— 3  ?/),  or  —2x  —  Sy,  is  also  its  square 
root ;  but  the  first  form  is  simpler,  and  will  be  used  in  all  the  examples 
of  the  present  chapter.) 

2.  Find  the  square  root  of  m^  —  2mn-\-  n^. 

By  the  rule,  the  result  is  m  —  n. 

(The  expression  may  also  be  written  w^  —  2  mn  +  m^  ;  in  which  case,  by 
the  rule,  its  square  root  is  w  —  m.) 

113.  Case  II.  When  the  expression  is  a  trinomial  perfect 
square.  y 

1.  Factor  25  a^  +  40  ab'  -^\      i'  a  H-  /->  ^ 
By  §  112,  the  square  root  of  the  expression  is  6  a  +  6*. 
Then,  25  a^  +  40  ah'^  +  &*  =  (5  a  ^62)2. 

2.  Factor  m^  —  4  m%^  +  4  n*. 


78  ALGEBRA 

By  §  112,  the  square  root  of  the  expression  is  either  w^  — 2n2,  oi 
2  n2  -  m2. 

Then,  m*  -  4  m'^n^  +  4  w^  =  (m2  -  2  n^y,  or  (2  w2  -  m^y. 

3.  Factor  a^- 2  %-2;)  +  (?/- 2)2. 
We  have  x^-2  x(y  -e)  +  (y-  0)2 

=  lx-iy-z)y=(ix-y  +  zyi 
or,  =[(y  -  0) -  a;]2  =  (2/  -  0  -  ic)2, 

4.  Factor  _  9  a*-  6  a^  - 1. 

-  9  a*  -  6  a2  _  1  =  _  ( 9  ^4  4.  6  a2  4. 1 )  =  _  (3  ^2  +  1 )  2. 

EXERCISE  35 

Factor  the  following: 

1.  q^  +  Sx  +  16.  5.   a^2/2  4-14a^  +  49. 

2.  9-6a  +  a'.  6.   36  0^-132  ab-^  121b' . 

3.  m2  +  10mn4-25w3^  7.    -16  a' +  24:  ax -9  a^. 

4.  4a«-4a36c2  +  6V.  8.   81  m^  + 180  m^  + 100  n^ 

^9.    -  25  a;io  -  60  x^fz'  -  36  ?/V. 
^10.   64:a'a^-2A0abxy  +  225by. 

11.  49  m^"  + 168  m*^^  + 144  a^^. 

12.  100  a2&2 -I- 180  a6c2  + 81  c^ 

13.  144a;V-312a^"2/^^  +  169«*». 
^14.    - 121  aV  + 220  a'b'mn- 100  b*n\ 

15.   169  a«62  + 364  a^6c2#4- 196  cW. 
^  16.   (a^  +  yf  +  22(0^  +  2/)  + 121.  ^; 

17.   a'-Sa(m-n)  +  16(m-nyj^ 
^18.   9a;2-6%  +  ^)  +  (2/  +  2)^ 

19.    (m  —  ?i)-  —  2(m  —  n)w  +  w^. 
'^20.   25(a  +  2>)'  +  40(a  +  6)c  +  16c2. 


FACTORING  79 

21.  36(a-aj)2-84(a-a;)2/  +  49/. 

22.  49  m^ -f- 42  m(m  + a;) +  9(m  +  a;)2. 
v^3.    (a  +  &)'  +  4(a  +  &)(a-&)  +  4(a-6)2. 

24.   9(a;  +  2/)'-12(a;4-2/)(a^-2/)+4(a;-2/)^ 

114.  Case  III.  When  the  expression  is  the  difference  of 
two  perfkct  squares.   ..,-, 

By  §98,  U\--b^=(a  +  b)(a-b). 

Hence,  to  obtain  the  factors,  we  reverse  the  rule  of  §  98 : 

Extract  the  square  root  of  the  first  square,  and  of  the  second 
square;  add  the  results  for  one  factor,  and  subtract  the  second 
result  from  the  first  for  the  other, 

1.  Factor  36a'b^-49c^ 

The  square  root  of  36  a^b^  is  6  aft^,  and  of  49  c^  is  7  c^. 
Then,  36  a^b^  -  49  c^  =  (6  ab^  +  7  c^)  (6  ab^  -  7  c^). 

2.  Factor  (2x-Syy-{x- y)\ 
By  the  rule,      (2  x  -  3  y)2  -  (x  -  y)'^ 

=  [(2x-32/)  +  (x-2/)][(2x-32,)-(x-2^)] 
=  (2x-3y  +  x-y)(2x-3y-x  +  y) 
=  (3x-4?/)(x-2?/). 

A  polynomial  of  more  than  two  terms  may  sometimes  be 
expressed  as  the  difference  of  two  perfect  squares,  and  factored 
by  the  rule  of  Case  III. 

3.  Factor  2  mri  +  m^  - 1  +  nl 

The  first,  second,  and  last  terms  may  be  grouped  together  in  the  order 
m2  +  2  TOW  +  n2  ;  which  expression,  by  §  112,  is  the  square  of  m  +  w. 

Thus,       2  TOn  +  to2  -  1  +  w2  =  (to2  +  2  mw  +  n2)  -  1 

=  (m  +  n)2-l 

=  (to  +  W  +  1)  (to  +  7i  —  1). 

4.  Factor  12y  +  ic2-9/-4. 


80  ALGEBRA 

122/  +  x2  -  9^2  _  4  =  a:2  -  9?/2  +  12 ?/  -  4 
=  a;2-(92/2-12  2/4-4) 
=:x2-(3?/-2)2,  by§112, 

=  (x  +  Sy-2)(x-Sy  +  2). 

5.   Factor  a2-c2  + 62  _c?2_  2  ccZ- 2  (i6. 
a2  -  c2  +  62  _  ^2  _  2  cd  -  2  a6 

=  a2  -  2  a6  +  62  _  c^^2cd-d^ 

=  a2  -  2  a6  +  62  -  (c2  +  2  cd  +  ^2)  =  (a  -  6)2-  (c  +  d)2,  by  §  112 

=  [(«-6)  +  (c  +  d)][(«-^>)-(c  +  «!)] 
=  (a  —  6  +  c  +  d)(a  —  6  —  c  —  d). 

EXERCISE  36 

Factor  the  following : 

1.  a^^m^\                    3.  n«-9.                     5.   l-64mV. 

2.  4a2-l.                  4.  16-25a«.              6.   36x'-121y\ 

7.  1-81  a«6V.  .  >/  19.  (ic  -  yf  -  (m  +  ^0'- 

8.  49a«-144  6V^.  20.  {2a  +  xy  -  {b -{-Syy. 

9.  100mV2-169nV«.  21.  (a  -  6)^  -  (c  -  d^. 

10.  225a.Y-196  2^  22.  (2m  +  ny -  (m-\-2ny. 

11.  324 a^™6io« _ 625.  23.  (6a-\-xy -  (a-Sxf. 

12.  36l£ci*-2562/^c^«.  24.  (9  a;  -  5  ?/)- -  (6  a;  -  7  ?/)'. 

13.  (a-by-c',~\  25.  25(8  a-3  6)2-9(4  a+5  6)2 

14.  (5x  +  yy-a^.l  26.  (a+6-c)=^-<a-6  +  <5)2. 
V15.  c4'*-(m  +  7i)l  }  27.  (w+w+3)2-(m-w-4)2. 

16.  a2-(6-2c)^\  28.  a' -^ 2 ab -\- b^ - c\ 

Vl7.  (x  +  4.v)2-922J  29.  x'-y^-2yz-z\ 

18.  36  m^  -  (2  m  -  3)2.  30.  m^- 7i2  +  2wp-i>2. 


FACTORING  81 

31,   a'-{-b'-l-2ab,  34.   9 a' -{- 16b'- 25 c' -{-24: ab. 

32/  y'  +  2xy-4.-\-x\  35.   9 -a' -i-2ab -b^ 

33/4:7)1^ —  4:mn-\-n'—p^.  36.   4:m' —p' —  9  n'  —  6np. 

37.   12yz  +  16x'-9z'-4y\ 
y38.   m^—2mn-{-n'  —  a:^  +  2xy  —  y\ 

39.  a'-{-2ab  +  b'-c'-2cd-d\ 
N/40.  a2  +  a:2_52_^2_j_2tta;  +  26?/. 
V/41.   x'-y'-\-7n^-l-2mx-2y. 

42.  a2-4aa;  +  4a^-&2_f.662/-92/2. 

43.  16a2_8a6  +  &'-c'-10cd-25d2. 
^44.   2Sxy-S6z'  +  49y''  +  60z-25  +  4x'. 

115.   Case  IY.     When  the  expression  is  in  the  form 

x'^-\-axhf-\-y^. 

Certain  trinomials  of  the  above  form  may  be  factored  by 
expressing  them  as  the  difference  of  two  perfect  squares,  and 
then  employing  §  114. 

1.  Factor  a* +  a262  +  6^ 

By  §  111,  a  trinomial  is  a  perfect  square  if  its  first  and  last  terms  are 
perfect  squares  and  positive,  and  its  second  term  plus  or  minus  twice  the 
product  of  their  square  roots. 

The  given  expression  can  be  made  a  perfect  square  by  adding  a%'^  to 
its  second  term ;  and  this  can  be  done  provided  we  subtract  a?h^  from  the 
result. 

Thus,  a*  +  a262  +  ^4  =  (^4  +  2  ^252  +  54)  _  ^252 

=  (a2  +  62)2  _  a%2^  by  §  112, 

=  (a2  _{.  52 ^  a6) (a2  +  ftf-  ah),  by  §  114, 

=  («2  +  a6  +  62)  (^2  _ah  +  62). 

2.  Factor  9  05*  -  37V  +  4. 

The  expression  will  be  a  perfect  square  if  its  second  term  is  — 12  x'^. 


82  ALGEBRA 

Thus,  9x*-37jc2-j-4  =  (9x*-  12x2  +  4)-25a;2 

=  (3  x2  -  2)2  -  (5  xy 

=  (3x2  +  5x-2)(3a;2_5aj_2). 

(The  expression  may  also  be  factored  as  follows : 

9  ic*  -  37  x2  +  4  =(9  x*  +  12  ^2  4-  4)  -  49  x'^ 

=  (3x2  +  2)2_(7x)2=(3.r2  +  7a:  +  2)(3x2-7a;+2) 

Several  expressions  in  Exercise  37  may  be  factored  in  two  different  ways. 
The  factoring  of  trinomials  of  the  form  x*  +  ax'^y^  +  y*,  when  the  factors 
involve  surds,  will  be  considered  in  §  300.) 

EXERCISE  37 

Factor  the  following : 
\   1.  .ic\+5a^  +  9.  ^^5.   9a;^  +  6a;y  +  49y*. 

2.  a^-21c&2_^36  6^  6.   16  a^ -  81  a^ -}- 16. 

3.  4  -  33  ar^  + 4  a;\  ^^1.   64-64m2  +  25m^ 

4.  25m*-14mV  +  w^  8.   49  a^  - 127  a V  +  81  iB*. 

Factor  each  of  the  following  in  two  different  ways  (compare 
Ex.  2,  §  115)  ; 

9.   a;^-17a^  +  16.  -11.   16  m*  - 104  mV  +  25  a;*. 

10.   9-148a2  4-64a^  12.   36a*-97aW  +  36m^ 

116.  Case  V.     When  the  expression  is  in  the  form 

x^  -\-  ax  +  b. 

We  saw,  in  §  99,  that  the  product  of  two  binomials  of  the 
form  X  -\-  m  and  x  -\-  n,  was  in  the  form  a^  +  aa;  -j-  & ;  where 
the  coefficient  of  x  was  the  algebraic  sum  of  the  second  terms 
of  the  binomials,  and  the  third  term  the  product  of  the  second 
terms  of  the  binomials. 

In  certain  cases,  it  is  possible  to  reverse  the  process,  and 
resolve  a  trinomial  of  the  form  x^  -\-  ax-{-h  into  two  binomial 
factors  of  the  form  x  -\-m  and  x  -\-n. 


\^  FACTORING  83 

To  obtain  the  second  terms  of  the  binomials,  we  simply  re- 
verse the  rule  of  §  99,  and  Jind  two  numbers  whose  algebraic  sum 
is  the  coefficient  of  x,  and  whose  product  is  the  last  term  of  the  tri- 
nomial. 

The  numbers  may  be  found  by  inspection. 

1.  Factor  a;2  4. 14  a;  4.  45. 

We  find  two  numbers  whose  sum  is  14  and  product  45. 

By  inspection,  we  determine  that  these  numbers  are  9  and  5. 

Whence,  a;2  +  14  x  +  45  =  (x  +  9)  (ic  +  5). 

2.  Factor  a^  —  5  ic  +  4. 

We  find  two  numbers  whose  sum  is  —  5  and  product  4. 
Since  the  sum  is  negative,  and  the  product  positive,  the  numbers  must 
both  be  negative. 

By  inspection,  we  determine  that  the  numbers  are  —  4  and  —  1. 

Whence,  a;2  _  5^;  +  4  =  (x  -  4)(x  -  1). 

3.  Factor  a?^  +  6  a;^  -  16. 

We  find  two  numbers  whose  sum  is  6  and  product  —  16. 

Since  the  sum  is  positive,  and  the  product  negative,  the  numbers  must 
be  of  opposite  sign  ;  and  the  positive  number  must  have  the  greater  abso- 
lute value. 

By  inspection,  we  determine  that  the  numbers  are  +  8  and  —  2. 

Whence,  x^ +  Qx^ -\Q  =  {y?  +  8)  {x^  -  2). 

4.  Factor  x""  -  abx"  -  42  a'b^. 

We  find  two  numbers  whose  sum  is  —  1  and  product  —  42. 
The  numbers  must  be  of  opposite  sign,  and  the  negative  number  must 
have  tlie  greater  absolute  value. 

By  inspection,  we  determine  that  the  numbers  are  —  7  and  +  6. 
Whence,      x^  -  abx^  -  42  ^252  _  (-^.2  _  7  ab)(c(^^  +  6  a?)). 

5.  Factor  l-^2a  —  99a\ 

We  find  two  numbers  whose  sum  is  +  2  and  product  —  99. 

By  inspection,  we  determine  that  the  numbers  are  +11  and  —  9. 

Whence,  1  +  2  a  -  99  a^  =  (1  +  11  o)  (1  -  9  a). 

If  the  x^  term  is  negative,  the  entire  expression  should  be 
enclosed  in  parentheses  preceded  by  a  —  sign. 


84  ALGEBRA 

6.   Factor  24^  +  5  x-x^. 

We  have,  24  +  5  a:  -  x^  =  -  (a;2  -  5  x  -  24) 

=  -(x-8)(x  +  3)  =  (8-a;)(3 +  x)  J 
changing  the  sign  of  each  term  of  the  first  factor. 

(In  case  the  numbers  are  large,  we  may  proceed  as  follows : 

Required  the  numbers  whose  sum  is  —  26  and  product  —  192. 
One  of  the  numbers  must  be  + ,  and  the  other  — . 
Taking  in  order,  beginning  with  the  factors  +  1  x  —  192,  all  possible 
pairs  of  factors  of  —  192,  of  which  one  is  +  and  the  other  — ,  we  have  : 

+  1  X  -  192. 

'  +  2  X  -    96. 

+  3x-    64. 

+  4  X  -    48. 

+  6  X  -    32. 

Since  the  sum  of  +  6  and  —  32  is  —  26,  they  are  the  numbers  required.) 

EXERCISE  38 

Factor  the  following : 

1.  x'-{-4,x  +  S.  13.  a^-17a;  +  52. 

2.  x'-Tx-hlO.  14.  a2  +  18a  +  56. 
.     3.  a^  +  Ta-lS.  ''  15.  S4:-{-5x-x\ 

4.  m^- 14  m -15.  16.  if-\-16y-57._ 

5.  2/2-162/  +  55.  17.  x'-lOx-TB. 

6.  x'  +  lQx  +  Sd.  18.  m2  +  19m  +  90. 
v7.  28  +  3c-c2.  19.  95-14  71-^2. 

8.  66-5n-nK  ^0.  x^- 20  a; +  96. 

9.  a2  - 14  a  +  48.  21.  a^  +  21  a  -f  98. 

10.  x^-\-20x-h51.  22.  x'-Tx-lS. 

11.  x^  - 12  X- 4:5.  23.  105-Sm-m'. 

12.  n2  +  14n-32.  24.  c^- 21  0^  +  104. 


FACTORING  85 

25.  x'-2Sx'-{-76.  43.  l-\-5a-UaK 

26.  a^'  +  a^-llO.  44.  m^  - 17  mn  +  66  nl 

27.  n''-16n'-S0.  45.  a^ +  12  ab +  27  b\ 

28.  a^"  + 18  a" +  65.  46.  a;^  -  14  ma?  +  40  m^. 

29.  cc2-  +  lla;"'-12.  47.  l-9x-36a;2. 

30.  c^-19c2^  +  88.  48.  m'  +  3mn-54:n\ 
v^31.  a^/-13a;?/3_3o;  49.  ^2  _^  12  a;^/ +  20  2/I 

32.  a'b^  -  23  ab'  + 112.  50.  a-V  - 17  a6c  +  60  c^. 

-   33.  A2  +  25  na;  4-154.  51.  l-lSn-6Sn\ 

'    34.  126  +  15?/'-/.  62.  a^  + 15  aa;  - 100  a^. 

'    35.  aV  +  9a-V-162.  53.  1  + 17  mw  +  70  mV. 

•  36.  m'^'*  -  23  77i3»  +  120.      '^       54.  a^^  - 17  a;^^^^  _^  72  i/V. 

•-  37.  (a  +  6)2  +  14(a  +  6)+24.     55.  a^  +  6  a'"'^  -  91  fe^. 

.    38.  {x-yy-15{x-y)-lQ.     56.  l-3xy-10^xY- 

39.  (m-n)2+21(m-r6)-130.  57.  a^ - 32  a6c  + 112  6V. 

^0.  (a  +  x)2-28(a  +  a;)  +  192.  58.  a^y  +  29  a^y^  - 170  ^l 

41.  a2_^6aa;  +  5a^.  59.  y-(2m  +  3  n)a;  + 6mn. 

42.  x'-Txy-Sy^  60.  x"- {a-b)x-ab. 

117.     Case  VI.    WJien  the  expression  is  in  the  form 

ax^  +bx  +  c. 

We  saw,  in  §  100,  that  the  product  of  two  binomials  of  the 
form  mx  +  n  and  px  +  q,  was  in  the  form  ax^  +  bx  +  c',  where 
the  first  term  was  the  product  of  the  first  terms  of  the  bino- 
mial factors,  and  the  last  term  the  product  of  the  second  terms. 

Also,  the  middle  term  was  the  sum  of  the  products  of  the 
terms,  in  the  binomial  factors,  connected  by  cross  lines. 

In  certain  cases  it  is  possible  to  resolve  a  trinomial  of  the 
form  ax^  +  bx  +  c  into  two  binomial  factors  of  the  form  mx  +  ?i 
and  px  +  q. 


86  ALGEBRA 

1.    Factor  3  a^  + 8  a; +  4. 

The  first  terms  of  the  binomial  factors  must  be  such  that  their  product 
is  3  a;2  .  ^^gy  can  be  only  3  x  and  x. 

The  second  terms  must  be  such  that  their  product  is  4. 

The  numbers  whose  product  is  4  are  4  and  1,-4  and  —1,2  and  2,  and 
—  2  and  —  2  ;  the  possible  cases  are  represented  below  : 


x  +  4 

x  +  1 

a;-4 

Sx  +  l 

3a;  +  4 

3a;-l 

13  X 

Ix 

-13  a; 

x-\ 

ic  +  2 

«-2 

3x-4 

3a;  +  2 

3x-2 

—  Ix  8ic  —  8a; 

The  corresponding  middle  term  of  the  trinomial,  obtained  by  cross- 
multiplication,  as  in  §  100,  is  given  in  each  case ;  and  only  the  factors 
X  +  2,  3  a;  +  2  satisfy  the  condition  that  the  middle  term  shall  be  8  x. 

Then,  3a;2  +  Sx  +  4  =  (x  +  2)(3a;  +  2). 

2.   ractor6a;2_^_2. 

The  first  terms  of  the  factors  must  be  6  x  and  x,  or  3  x  and  2  x. 
The  second  terms  must  be  2  and  —  1,  or  —  2  and  1. 
The  possible  cases  are  given  below  : 


6x  +  2 

6x-l 

6x-2 

6x  +  l 

x-1 

x  +  2 

x  +  1 

x-2 

-4x 

11  X 

4x 

-llx 

3x  +  2 

3x-l 

3x-2 

3x  +  l 

2x-l 

2x  +  2 

2x+l 

2x-2 

X  4x  —X  — 4x 

Only  the  factors  3  x  —  2  and  2  x  +  1  satisfy  the  condition  that  the 
middle  term  shall  be  —  x. 

Then,  6x2 -x  -  2  =  (3x  -  2)(2x  +  1). 

The  following  suggestions  will  be  found  of  service : 

(a)  If  the  last  term  of  the  trinomial  is  positive,  the  last  terms 
of  the  factors  will  be  both  -f,  or  both  —,  according  as  the  middle 
term  of  the  trinomial  is  +  or  — . 


FACTORING  87 

Thus,  in  Ex.  1,  we  need  not  have  tried  the  numbers  —  1  and  —  4,  nor 
—  2  and  —  2 ;  this  would  have  left  only  three  cases  to  consider. 

(b)  If  the  last  term  of  the  trinomial  is  negativey  the  last  terms 
of  the  factors  will  be  one  -\~,  the  other  — . 

If  the  X-  term  is  negative,  the  entire  expression  should  be  enclosed  in 
parentheses  preceded  by  a  —  sign. 

If  the  coefficient  of  a:^  is  a  perfect  square,  and  the  coefficient 
of  X  divisible  by  the  square  root  of  the  coefficient  of  x^,  the  ex- 
pression may  be  readily  factored  by  the  method  of  §  116. 

3.   Factor  9  a;2  _  18  a;  +  5. 

In  this  case,  18  is  divisible  by  the  square  root  of  9. 

We  have  9  a:2  -  18  x  +  5  =  (3  x)2  -  6(3  x)  +  6. 

We  find  two  numbers  whose  sum  is  —  6,  and  product  5. 
The  numbers  are  —  5  and  —  1. 

Then,  9x2  -  ISx  +  5  =  (3a:  -  5)(3a;  -  1). 

^EXERCISE  3#  /  >         3 

Factor  the  following : 

VI.    2,x'  +  9x-\-9.  12.  10a^-39a;  +  14. 

:>2.   3ar2-lla;-20.  >1S.  12  a;^  + 11  a;  +  2. 

3.  4a:2-28a;  +  45.  14.  20  a V - 23  aa; -f  6. 

4.  ex'-hTx-S,  n5.  36a;2-|-12a;-35. 

5.  5a;2-36^'  +  36.  16.  6-aj-15a^. 

6.  16a;2  +  56a;  +  33.  17.  5  +  9x-18ar^. 

7.  8n2  +  18n-5.  18.  72H-7a;-49x^ 

8.  4.x^-Sx-7.  >19.  24a;2-17wa;  +  3n2. 

9.  9a;2  +  12a;-32.  20.  28a^-a;-2. 

-10.   6x'  +  7ax-^2a^.  21.   21  a^"»  +  23  a; V  +  ^  ^^ 

11.   25x'-25mx-6m\'  22.    18  a^  -  27  afta;  -  35  r/6-. 

v23-   24  tt^  +  26  tt^- 5. 


88  ALGEBRA 

118.  It  is  not  possible  to  factor  every  expression  of  the  form 
a^  +  aa;  H-  6  by  the  method  of  §  116. 

Thus,  let  it  be  required  to  factor  a;^  +  18  a;  +  35. 

We  must  find  two  numbers  whose  sum  is  18,  and  product  35. 

The  only  pairs  of  positive  integral  factors  of  35  are  7  and  5, 
and  35  and  1 ;  and  in  neither  case  is  the  sum  18. 

It  is  also  impossible  to  factor  every  expression  of  the  form 
axr -\-hx-\-c  by  the  method  of  §  117. 

Thus,  it  is  impossible  to  find  two  binomial  factors  of  the 
expression  4  a;^  +  4  a;  —  1  by  the  method  of  §  117. 

In  §  298  will  be  given  a  general  method  for  the  factoring  of 
any  expression  of  the  forms  x^ -\-ax  -\-h,  or  ax^  -\-hx-\-c. 

119.  If  an  expression  can  be  "resolved  into  three  equal  fac- 
tors, it  is  said  to  be  a  perject  cube,  and  one  of  the  equal  factors 
is  called  its  cube  root. 

Thus,  since  27  o!^W  is  equal  to  3  a-6  x  3  a-h  x  3  o?h,  it  is  a 
perfect  cube,  and  3  a^h  is  its  cube  root. 

120.  The  following  rule  for  extracting  the  cube  root  of  a 
positive  monomial  perfect  cube  is  evident  from  §  119 : 

Extract  the  cube  root  of  the  numerical  coefficient^  and  divide 
the  exponent  of  each  letter  by  3. 

Thus,  the  cube  root  of  125  a^6V  is  5  a^bh. 

121.  Case  YII.  When  the  expression  is  the  sum  or  difference 
of  two  perfect  cubes. 

By  §  102,  the  sum  or  difference  of  two  perfect  cubes  is  divis- 
ible by  the  sum  or  difference,  respectively,  of  their  cube  roots ; 
in  either  case  the  quotient  may  be  obtained  by  the  rules  of  §  102. 

1.   Factor  a;^- 27  2/V. 

By  §  120,  the  cube  root  of  x^  is  x^,  and  of  27  yV  is  3  yH. 

Then  one  factor  is  x^  —  3  y^z. 

Dividing  a^  —  27  y'^z'^  by  a;^  —  3  y^z.,  the  quotient  is 

X*  +  3  xhiH  +  9  ?/62-2  (§  102). 
Then,  x^  -  27  y^z^  =  (x^  _  3  yH)  (cc*  4-  3  x^yH  +  9  fz'^). 


^^V_^^^; 


FACTORING  89 

2.  Factor  a«  4- &^ 
One  factor  is  a^  4.  ^2^ 

Dividing  a^  +  h^  by  a^  +  &'^,  the  quotient  is  a*  —  a^ft^  +  54, 
Then,  a^  +  56  ^  (^2  +  52)  (a*  _  a25-2  +  ^4). 

3.  Factor  {x  +  a)'''  —  {x  —  of. 
(x  +  a)3-(x-a)3 

=  [(cc  +  a)  -  (x-a)][(x  +  a)2+  (a;+a)(a;-a)  +  (x-a)2] 
=  (ic  +  a  -  ic  +  a)  (x2  +  2  ax  +  a2  +  x2  -  a2  +  x2  -  2  ax  +  a2) 


—  -"  "<  V"  •*'  T^  "*  < 

!• 

EXERCISE  ^ 

^0 

a.  -      cx-lr- 

Factor  the  following : 

1.   a?  +  b\ 

i^v 

Sa^  +  l. 

9.   64m3  — n^ 

2.   ^-y\ 

.6. 

l-27nl 

vlO.   a«6«-216cl 

3.   l  +  m«n^ 

7. 

a«  +  l. 

11.    8m3^  +  27n^. 

4.   a^-ft^c^. 

8. 

ccy  +  z«. 

12.   27iB«--125/«. 

13.   64  +  125a365. 

18. 

729a«6«  +  8c^dl 

14.   343a«-64m^ 

vl9. 

(^ 

4_2/)3  +  (a;-2/)3. 

15.   125a^+216/;2^ 

20. 

m^ 

—  (m  +  w)^. 

16.   27a«-512  6». 

21. 

27 

(a-6)3  +  8  6^ 

17.   216aV-343n« 

1 

22. 

(2 

a  +  a^)^— (a  +  2a;)^ 

1/23.    {hx-2yf-{?>x-^yf. 

122.  Case  VIII.  TF7ie?i  the  expression  is  the  sum  or  differ- 
ence of  two  equal  odd  powers  of  t2vo  numbers. 

By  §  104,  the  sum  or  difference  of  two  equal  odd  powers  of 
two  numbers  is  divisible  by  the  sum  or  difference,  respectively, 
of  the  numbers. 

The  quotient  may  be  obtained  by  the  laws  of  §  103. 

Ex.   Factor  a^  +  6^ 

By  §  104,  one  factor  is  a  +  &. 


90  -'^-  ALGEBRA 

(/■ 

Dividing  a^  +  ft^  by  a  +  &,  the  quotient  is 

a*  -  a%  +  a2&2  _  ah^  +  6*.  (§  103) 

Then,         (a^  +  6^)  ^  («  +  5)  (^4  _  ^85  _f.  ^252  _  ^h^  _j.  54). 

EXERCISE  41 

Factor  the  following : 

vl.   ar'  +  Z.                    5.   1+a;^  9.  32a«-6«. 

2.  a^-1.                   v6.   a;^  +  7i9^  >10.  243a^  +  2/^. 

3.  1-mV.                 7.   a«-l.  v^  11.  m"  +  128n^ 

4.  a7-6^                     8.   n«^  +  32.  v  12.  32  a^^^^n  _  243  ciop. 

123.  By  application  of  the  rules  already  given,  an  expres- 
sion may  often  be  resolved  into  more  than  two  factors. 

If  the  terms  of  the  expression  have  a  common  factor,  the 
method  of  §  107  should  always  be  applied  first. 


f 


1.  Factor  2  aa^if  —  8  axy^. 

By  §  107,    2  ax^y^  -  8  axy^  =  2  axy^x"^  -  4y^) 

=  2  axy\x  +  2  ?/)  (x  -  2  y),  by  §  114. 

2.  Factor  .a«-5^ 

By  §  114,  a6  -  &6  =  (a^  +  b^)  (a^  -  63). 

Whence,  by  §  121, 

a6-b^  =  (a-\-  h){a^  -  ah  +  h'^){a  -  h){aP-  +  a6  +  62). 

3.  Factor  ajs-/. 

By  §  114,  a^  -  ^  =  (X*  +  y4)(x4  -  y^) 

=  (a;*  +  t){'^^  +  y^){^  +  y){^  -  2/). 

4.  Factor  3  (m  +  w)^  _  2  (m^  -  n^). 

3(m  +  w)2  -  2(m2  -  rfi)  =  3(m  +  n)2  -  2(m  +  w)(wi  -  w) 
=  (m4- w)[3(w  +  n)-2(m-n)] 
=  (m  +  w)  (3  ?n  4-  3  n  -  2  m  +  2  w) 
=  (w  +  n)(»i  +  bn). 


FACTORING  91 

5.   Factor  a(a-l)- 6  (6-1). 

a(a  -  1)  -  6(6  -  1)  =  a2  -  a  -  62  +  5 
=  a2  _  62  _  05  +  5 
=  (a  +  6)(a-6)-(a-6) 
=  (a-6)(a  +  6-l). 


The  following  is  a  list  of  the  type-forms  in  factoring,  con- 
sidered in  the  present  chapter : 


ax  -\-  ay  —  az. 

(§  107) 

jr^  +  fljr  +  b. 

(§  116) 

fl-  +  2  a6  4-  61 

(§  113) 

ax^  -^  bx  +  c. 

(§  117) 

fl^  -  2  a6  +  b\ 

(§  113) 

a'  +  b\ 

(§  121) 

a'  -  b\ 

(§  114) 

a'  -  b\ 

(§  121) 

x'  +  axy  +/. 

(§  115) 

a"  -  b\ 

(§  122) 

a"  +  6%  n  odd.     (§  122) 

MISCELLANEOUS  AND  REVIEW  EXAMPLES 
EXERCISE  42 
Factor  the  following : 

1.  0^24.210^  +  98,  4.   49aV-210a26«c  +  225  6'\ 

2.  a2-20a-69.  5.   21a^ -{-Uo}-24.a-lQ. 

3.  12aj«-18a5^  +  6a;^-9a^.     76.   a' - 22 a^ -\- 120. 

>7.   49  a;2_|_  49^.^12. 
8.   7a;2(3a-26)-3a^(2a-36). 

9.   x'-16.  15,  45n«  +  18n2-207i-8. 

10.   24 a6 -18 a?/ -20 6a; +15 0^2/.   16.  96-20a-a2. 

^11.   27a3  +  1000.  -A7.  36  a;^  -  69  a;^  _|_  25. 

12.  646  m'- 250 mril  18.  2a'x-Sa'a:^-}-2a^afi-Saxl 

13.  (x'-{-x-2y-(x'-x-\-3y.   19.  128a;y+288a;y+162a;/. 

14.  64: -n\  20.  2  xi/ -  2  ic^^^  _  254  a;^^^. 


92  ALGEBRA 

21.  m'^-l.  24.    -121m«  +  22m*-l. 

22.  a^'-l.  25.   36x^-\-24.a^-21x\ 

23.  a%'- SO  a^bc^-\- 216  c\  26.   a'b^  +  aY  -  ¥x^  -  x'f. 

27.  (a-\- 2  by  +  S(a-^  2b){2  a-b) +16(2  a -by, 

28.  4a;(a-&-c)  +  02/(6  +  c-a). 

29.  (m  +  7i)*-2(m  +  ^)^4-(m  +  w)^. 

30.  x^-16xy-^64:y^  32.   a;6-26^-27. 

31.  Slm'-256n^  33.    (a;  +  2  2/)^  +  (3  x  -  2/)^ 

34.   (a  +  2a;)2  +  10(a  +  2x)-144. 

35.  27 x''-75y^-120yz-ASz\  39.   49 a^^^ _f_  12 a^^e _^ 4 a^^^^ 

36.  (a' -{-4.ab  +  b')'-(a' -j-b'y.  40.   14 a;^ _ 25 ^^ 4. 6. 

37.  (16m"  +  n2)2-64mV.  41.   a''-x'\ 

38.  49tt2  +  4-3662_28a.         42.   x^'-2x'  +  l. 

43.  9  aV  - 16  a'd'  -  36  6V  +  64  bH\ 

44.  mV-243mV.  46.   a'  +  12Sb\ 

45.  _7aj2-26a;  +  8.  47.   4S x^y -  52 xY -  ^^0 xy\ 

48.  Kesolve  a^  —  81  into  two  factors,  one  of  which  is  a  —  3. 

49.  Eesolve  of  —  64  into  two  factors,  one  of  which  is  x-\-2. 

50.  Eesolve  x^  —  y^  into  two  factors,  one  of  which  is  x  —  y. 

51.  Eesolve  a^  +  1  into  two  factors,  one  of  which  is  a  + 1- 

52.  Eesolve   l-\-x^  into  three   factors   by  the   method  of 
Case  VII. 

53.  Eesolve  a^  —  512  into  three  factors. 

Factor  the  following : 

54.  a^  —  m^ -\- a -\- m. 

55.  (x2-f  4a?)2-37(ar^  +  4x)  +  160. 

56.  iii*'-1024.  57.   m^  +  m  +  x^-i-x. 


FACTORING  93 

58.  a^(^  -  4.  b'c^  -  S  a'd^-\- 32  b'd^ 

59.  (m-7i)(aj2-2/2)  +  («  +  ^)(m2-w2). 

60.  (x-iy  +  6(x-iy  +  9(x-l). 

61.  (m  +  n)  (771^  —  a^)  —  (m  +  a?)  (m^  —  n^. 

62.  a2-4&2_a-26.  63.    (a;^  +  4 2/' - i^y - 16 a;y . 

64.  {x^ -9xy-{-4.(x^-9x)- 140. 

65.  d'b'-h27ay-Sb^a^-2Wa^f. 

66.  (m^  +  m)2  +  2  (m^  +  m)(m  + 1)  +  (m  + 1)'. 

67.  (2a;2_3)2-ar^.  69.    (4:  a' -  b' -  9y  -  36  b\ 

68.  64aV  +  8a3_8ar'-l.       70.    (x  +  2yy-x(x'-4:y^). 

71.  16aj2_|.2/2_25;32_i^83.^_j_10^. 

72.  (a2  +  6a  +  8)2-14(a2  +  6a  +  8)-15. 

73.  (l  +  ar^)  +  (l+a;)3.  75.    (a^  + /)  _  a;^/ (x  +  2/). 

74.  a*-9  +  2a(a2  +  3).       '      76.    (a^- 8  m«)-a(a-2m)2. 

77.   9a2(3a  +  2)2  +  6a(3a  +  2)  +  l. 

78.  m»-m^  +  32m3-32.  80.   m\m-^p) +n^(n-p).  ^ 

79.  a(a-c)-6(6-c).  81.   a;^  +  8 a;«  +  a^-"  +  8. 

82.  (27m^-a^)  +  (3m-\-x){9'm'-6mx-{-x^. 

83.  (4a2  +  9)2_24a(4a2  +  9)  +  144a2. 

84.  16a2  +  962_25c2-4d2-24a6-20cd 

85.  m9  +  m^-64m«-64. 

86.  (a;2^2/0'-4a;y(aj2  4-2/2). 

87.  a'-\-a''b-{-a^b^-{-a^b^-{-ab^-{-b\ 

88.  (87i3_27)+(27i-3)(4n2 4-471-6). 

89.  a^  +  2ar^  +  2a;  +  l. 

(By  altering  the  order  of  the  terms,  this  may  be  written 

x8+l  +  (2a;2  +  2x),  or  (x  +  l)(x'^  -  x  +  1) -\- 2x  {x-\- 1), 
and  X  +  1  is  a  factor  of  the  given  expression.) 


94  ALGEBRA 

90.   x^-3x''-\-3x-l. 

92.   S  c^  +  36 x'y  +  5Axy' -}- 27 if . 

Additional  methods  in  factoring  will  be  found  in  §§  298  to 
300,  and  in  Chapter  XXXIV. 

124.  By  §  54,  (+  a)  x  (+  6)  =  +  ab,  (+  a)x(-b)  =  -  ab, 

{-a)x(+b)=-ab,  {- a)x{- b)  =  + ab. 

Hence,  in  the  indicated  product  of  two  factors,  the  signs  of 
both  factors  may  be  changed  without  altering  the  product;  but  if 
the  sign  of  either  one  be  changed,  the  sign  of  the  product  will  be 
changed. 

If  either  factor  is  a  polynomial,  care  must  be  taken,  on 
changing  its  sign,  to  change  the  sign  of  each  of  its  terms. 

Thus,  the  result  of  Ex.  3,  §  107,  may  be  written  in  the  forms 

(b  —  a)(n  —  m),  —  (b  —  a)(m  —  n),  or  —  (a  —  b)(n  —  m). 

In  like  manner,  in  the  indicated  product  of  more  than  two 
factors,  the  signs  of  any  even  number  of  them  7nay  be  changed 
without  altering  the  product;  but  if  the  signs  of  any  odd  number 
of  them  be  changed,  the  sign  of  the  product  will  be  changed  (§  b5). 

Thus,  {a  —  b){c  —  d)(e  — /)  may  be  written  in  the  forms 
(a-bXd-c)(f-e), 
(b-a)(c-d)(f-e), 
—  (6  —  a)(d  —  c)(/—  e),  etc. 

SOLUTION  OF  EQUATIONS  BY  FACTORING 

125.  Let  it  be  required  to  solve  the  equation 

(a;-3)(2a;  +  5)  =  0. 

It  is  evident  that  the  equation  will  be  satisfied  when  x  has 
such  a  value  that  one  of  the  factors  of  the  first  member  is 
equal  to  zero ;  for  if  any  factor  of  a  product  is  equal  to  zero, 
the  product  is  equal  to  zero. 


FACTORING  96 

Hence,  the  equation  will  be  satisfied  when  x  has  such  a  value 
that  either  aj_3  =  0,  (1) 

or  2x  +  5  =  0.  (2) 

5 
Solving  (1)  and  (2),  we  have  a;  =  3  or  —  -• 

It  will  be  observed  that  the  roots  are  obtained  by  placing  the 
factors  of  the  first  member  separately  equal  to  zero,  and  solving 
the  resulting  equations. 

126.   Examples.  "^ 

1 .  Solve  the  equation  a^  —  5  a;  —  24  =  0. 

Factoring  the  first  member,  (x  -  8)  (x  +  3)  =  0.  (§  116) 

Placing  the  factors  separately  equal  to  0  (§  125),  we  have 

X  —  8  =  0,  whence  x  =  8  ; 

and  X  +  3  =  0,  whence  x  =  —  3. 

2.  Solve  the  equation  4  a;^  —  2  a;  =  0. 

Factoring  the  first  member,  2  x(2  x  —  1)  =  0. 
Placing  the  factors  separately  equal  to  0,  we  have 

2  X  =  0,  whence  x  =  0  ; 

and  2  x.—  1=0,  whence  x  =  -  • 

2 

3.  Solve  the  equation  ar'  +  4a^  —  a;  —  4  =  0. 
Factoring  the  first  member,  we  have  by  §  108, 

(x  +  4)  (x2  -  1)  =  0,  or  (X  +  4)  (X  +  1)  (x  -1)  =  0. 
Then,  x  +  4  =  0,  whence  x  =  —  4  ; 

X  +  1  =  0,  whence  x  =  —  1 ; 
and  X  —  1  =  0,  whence  x  =  1. 

4.  Solve  the  equation  a^  _  27  -  (a^  +  9  a;  -  36)  ==  0. 
Factoring  the  first  member,  we  have  by  §§  116  and  121, 

(x  -  3)(x2  +  3x  +  9)  -  (X  -  3)(x  +  12)  =  0. 
Or,  (x-3)(x2  +  3x  +  9-x-12)  =  0. 

Or,  (x-3)(x2  +  2x-3)=0. 


96  ALGEBRA 

Or,  (x-3)(a;  +  3)(x-l)  =  0. 

Placing  the  factors  separately  equal  to  0,  x  =  3,   —  3,  or  1. 
The  pupil  should  endeavor  to  put  down  the  values  of  x  without  actu- 
ally placing  the  factors  equal  to  0,  as  shown  in  Ex.  4. 

EXERCISE  43 

Solve  the  following  equations : 

1.  aj2_^7a;  =  0.  '  11.  a;^  + 18  a^  +  32  a;^  ^  0^ 

2.  5a3-4a^  =  0.     '^   ^  12.  x' -13x'  +  S6  =  0. 

3.  3a^-108ic  =  0.  13.  8  a;2_i0a;  +  3=:0. 

4.  (3ic-2)(4a;^-25)  =  0.  14.  6  x' +  7  x +  2  =  0. 

5    x^ -15 X  + 54:  =  0.  15.  3x'-mx-4:m^  =  0. 

6.  a^  +  23a;  +  102  =  0.  16.  lO-x^-f  7  a;-12  =  0. 

7.  x2  +  4a;-96  =  0.    "  17.  15x^+x-2  =  0. 

8.  aj^-x-110  =  0.    ^    '  18.  12a;3-29aj2  +  15a;  =  0. 

9.  x'-^ax-2a^  =  0.                    19.  x" -ax-{-bx-ab  =  0. 
10.   (5a;  +  l)(a^-6ic-91)  =  0.     20.  x^  +  mx -\- nx -{- mn  =  0. 

21.  a^-2cx-8aj  +  16c  =  0. 

22.  aj^  +  3  m-a;  — Sm^a;— 15m^  =  0. 

23.  27a^  +  18x2_3^_2  =  o. 

24.  (aj-2)2-4(a^-2)+3  =  0. 

25.  (4  a;2 -  49)(a;2  _  3  ^  _  lo^)(3  ^,2  _^  ^4  ^^  _  15)  ^  q 

26.  (a:-2)(5a;2  +  3a;-4)-(a^-4)  =  0. 

27.  (aj2-l)(a^-9)  +  3(a;-l)(a;  +  3)  =  0. 


HIGHEST   COMMON   FACTOK  97 


IX.     HIGHEST    COMMON    FACTOR.     LOWEST 
COMMON  MULTIPLE 

(We  consider  in  the  present  chapter  the  Highest  Common  Factor  and 
Lowest  Common  Multiple  of  Monomials^  or  of  polynomials  which  can 
be  readily  factored  by  inspection. 

The  Highest  Common  Factor  and  Lowest  Common  Multiple  of  poly- 
nomials which  cannot  be  readily  factored  by  inspection,  are  considered 
in  §§  439  to  443.) 

HIGHEST  COMMON  FACTOR 

127.  The  Highest  Common  Factor  (H.  C.  F.)  of  two  or  more 
expressions  is  their  common  factor  of  highest  degree  (§  64). 

If  several  common  factors  are  of  equally  high  degree,  it  is  understood 
that  the  highest  common  factor  is  the  one  having  the  numerical  coefficient 
of  greatest  absolute  value  in  its  term  of  highest  degree. 

For  example,  if  the  common  factors  v^ere  6  aj  —  3  and  2  a:  —  1,  the 
former  would  be  the  H.  C.  F. 

128.  Two  expressions  are  said  to  be  prime  to  each  other  when 
unity  is  their  highest  common  factor. 

129.  Case  I.     Highest  Common  Factor  of  Monomials. 

Ex,     Required  the  H.  C.  F.  of  42  a?h\  70  d'hc,  and  98  a'hH\ 

By  the  rule  of  Arithmetic,  the  H.  C.  F.  of  42,  70,  and  98  is  14. 
It  is  evident  by  inspection  that  the  expression  of  highest  degree  which 
will  exactly  divide  a%'^,  a%c,  and  a%hP  is  a%. 

Then,  the  H.  C.  F.  of  the  given  expressions  is  14  a^h. 

It  will  be  observed,  in  the  above  result,  that  the  exponent  of 
each  letter  is  the  loivest  exponent  with  which  it  occurs  in  any  of  the 
given  expressions. 

EXERCISE  44 

Find  the  H.  C.  F.  of  the  following: 

1.   14a^2/,  21xy\  2.   Ua'b\  112  6V. 


98  ALGEBRA 

3.  60(x-yy,  S4:(x-yy.  5.    72  a'b^,  27a%^,  99  a'b*. 

4.  108  m^ny,  90  7nhipl  6.    Ux'yz%  SSxYz',  110  ar^T/'^ 

7.  32a'x\  128  aWa^,  192  a^Y- 

8.  136  a%V,  516%)i«,  119  c^??^^. 

9.  72xyz',  168a^?/V,  120ajV2j-r 

10.   26(a  J6)2(a-6/,  91{a  +  by{a-by. 


130.   Case  II.     Highest  Common  Factor  of  Polynomials  which 
can  be  readily  factored  by  Inspection. 

1.  Kequired  the  H.  C.  F.  of 

5  x^y  -  45  T'y  and  10  a^/  -  40  xY  -  210  xy^ 

By  §§  107,  114,  and  116,  5  x^y  -  45x^y  =  6x^y(x^  -  9) 

=  5x^y(ix  +  3)(x-S);  (1) 

and  10  x^y^  -  40  xV  _  210  xy^  =  10  xy'^ix'^  -  4  x  -  21) 

=  10xy\x-7)(x-hS).         (2) 

The  H.  C.  F.  of  the  numerical  coefficients  5  and  10  is  5. 
It  is  evident  by  inspection  that  the  H.  C.  F.  of  the  literal  portions  of  the 
expressions  (1)  and  (2)  is  xy(x  +  3). 

Then,  the  H.  C.  F.  of  the  given  expressions  is  5xy{x  +  3). 

It  is  sometimes  necessary  to  change  the  form  of  the  factors  in 
finding  the  H.  C.  F.  of  expressions. 

2.  FindtheH.C.F.  of  a2  +  2a-3  and  1-al 
By  §116,  a2  +  2a-3  =  (a-l)(a  +  3). 

By  §121,  l-a^  =  (l-a){l  +  a  +  a^). 

By  §  124,  the  factors  of  the  first  expression  can  be  put  in  the  form 

-(l-a)(3  +  a). 
Hence,  the  H.  C.  F.  is  1  -  a. 

EXERCISE  45 
Find  the  H.  C.  F.  of  the  following  : 

1.   SOicy  +  lOary,  15a;y-30a;/. 


HIGHEST  COMMON   FACTOR  99 

2.  a'-ieb',  a'  +  Sab-\-16b\ 

3.  m"-14m  +  45,  m'-10m-j-25. 

5.  a3  +  64,  a'-7a-A4.. 

6.  9-a^,  aj2-aj-6. 

7.  ac  —  6c  —  ad  +  bd,  d^  —  cl 

8.  a;2 -f  13  a;  +  22,  2  aj^  +  9  a;  + 10. 

9.  3ac-4ad-66c  +  86d,  a-H-7a&-1862. 

10.  x^  +  y''-z'^-2xy,  a? -y'' -z" +  2yz. 

11.  3ar'-16a'?/  +  5  2/^  a;^  +  10 a;?/ - 75 /. 

12.  m^-^m\  m^  +  4m2  +  16. 

13.  2a^-7ajH-6,  6aj2-llx  +  3. 

14.  2ar'-13a;i/  +  62/S  xy^-^o?. 

15.  l-lla  +  18a2,  8a«-l,  18a2-5a-2. 

16.  Sa?-2Qo?b-\-20ab\  12  a^ -10  a^b- 2^  db\ 

17.  a;2  +  18a;  +  77,  a;^  +  22 a;  + 121,  a^  +  a^-HO. 

18.  16m2-9n2,  16m2-24mn +  9  n^  9  mn^  -  12  rri^n. 

19.  ar^-27,  a^-6a;  +  9,  2  aa;-6  a- 6a;  +  3  6. 

20.  27a3  +  863,  9a2-46^  9aH12a6  +  462. 

21.  a2-3a-18,  2a2-a-21,  3a2  +  4a-15. 

22.  2a^-12a^  +  16aj,  3 a;* - 3 a^ - 36 ;t^,  5 a:^ 4- 5 a;^ -  100 a:^^ 

23.  125m*-8m,  10 m^ -\- m^ - 2 m,  25 m^ - 20 rrv' -{- ^ m. 

24.  a^  +  3a2-40,  a*-2o,  a'-^a'-5a-5. 

25.  2a:3_aj2_g3,_j_3^  6.T2-19a;  +  8,  4a;2_^3^__5^ 

26.  a'-(b  +  cy,  {b-ay-(^,  b'-(a-cy. 

27.  Sx^y-\-xY,  64:xy-{-2xf,  24:  a^y  -  SO  aPy^  -  21  xy^.  y 

28.  2a'^  +  17a4-36,  4a2-4a-99,  6a2  +  25a- 9.         '"'^'^ 


100  ALGEBRA 

LOWEST    COMMON  MULTIPLE 

131.  A  Common  Multiple  of  two  or  more  expressions  is  an 
expression  which  is  exactly  divisible  by  each  of  them. 

132.  The  Lowest  Common  Multiple  (L.  C.  M.)  of  two  or  more 
expressions  is  their  common  multiple  of  lowest  degree. 

If  several  common  multiples  are  of  equally  low  degree,  it  is  understood 
that  the  lowest  common  multiple  is  the  one  having  the  numerical  coeffi- 
cient of  least  absolute  value  in  its  term  of  highest  degree. 

For  example,  if  the  common  multiples  were  4  a:— 2  and  6x  — 3,  the 
former  would  be  the  L.  C.  M. 

133.  Case  I.     Lowest  Common  Multiple  of  Monomials. 

Ex.   Required  the  L.  C.  M.  of  36  a%  60  aV,  and  84  cv?. 

By  the  rule  of  Arithmetic,  the  L.  C.  M.  of  36,  60,  and  84  is  1260. 
It  is  evident  by  inspection  that  the  expression  of  lowest  degree  which 
is  exactly  divisible  by  a%,  aP-y'^,  and  cx^  is  a^cx^y'^. 

Then,  the  L.  C.  M.  of  the  given  expressions  is  1260  aHx^y^. 

It  will  be  observed,  in  the  above  result,  that  the  exponent  of 
each  letter  is  the  highest  exponent  with  which  it  occurs  in  any  of 
the  given  expressions. 

EXERCISE  46 

Find  the  L.  C.  M.  of  the  following : 

1.  6  a^f,  6  xy.  5.   105  a%  70  b'c,  63  c'a. 

2.  lSa%4:5b'c.  6.   50  xy,  2i  a^f,  4:0  a^y\ 

3.  2Sa^,36y\  7.   21  ab*,  35  b'c%  91  a'(^. 

4.  42  m%^  98  ny.  8.   56  a'b^,  84  ba^,  48  xy. 

9.   60  a'bc',  75  a'b%  90  aVcZ«. 
10.   99  m^nx«,  66  m^nV,  165  wV/. 

134.  Case  II.  Lowest  Common  Multiple  of  Polynomials  which 
can  be  readily  factored  by  Inspection. 

1.    Kequired  the  L.  C.  M.  of 

ar^  —  5a;  +  6,  ar^  —  4a;-|-4,  and  a^  —  9  x. 


LOWEST   COMMON  MULTIPLE  101 

By  §  116,  a:2  -  5  x  +  6  =  (X  -  3)  (a;  -  2). 

By  §  113,  x2-4x  +  4=(x-  2)2. 

By  §  114,  x8-9x  =  x(x  +  3)(x-3). 

It  is  evident  by  inspection  that  the  L.  C.  M.  of  these  expressions  is 

x(x-2)2(x  +  3)(x-3). 
It  is  sometimes  necessary  to  change  the  form  of  the  factors. 

2.    Find  the  L.  C.  M.  of  ac-bc-ad-^  bd  and  b-  -  a^. 

By  §  108,   ac-hc-ad^hd  =  {a-  b)  (c-  d). 

By  §114,  62-a2=  (&  +  a)(&-a). 

By  §  124,  the  factors  of  the  first  expression  can  be  written 

\b-a)id-c). 
Hence,  the  L.  C.  M.  is  (6  +  a)  (6  -  a)  (d-c),  or  (62  _  a^)  (d-c). 

EXERCISE  47 
Find  the  L.  C.  M.  of  the  following : 

2.  o?b-\-2a^b\2a^b''-{-ab\ 

3.  m2-6m  +  9,  m2-llm  +  24. 

4.  a*  -  49  a'bS  a'  + 12  a^b  +  35  o?b\ 

5.  2a^  +  2a;2-84a;,  3x3-3a^-90aj. 

6.  a^-o^,a^-a?x-\-ao?-o?. 

7.  1  +  27  0.-3,  l-5a;-24aj2. 

8.  ac  —  3ad-2bc  +  Q>bd,^ac  +  ad  —  Q>bc  —  2bd. 


10.  a2_73a  +  10,  10-5a  +  2a2-a^  -  l^     \^ ^    ^ 

11.  a^  +  8,  4a;2-(a^  +  4)2.  vi         "^^ 

12.  2a^  +  3a;-35,  2ar^  +  19a:  +  45. 

13.  9n--27n  +  8,  3rj2-2n-16.       .  ,  . 


102  ALGEBRA 

14.  16x'-25y',12x^-\-15xy,Sxy-10y\ 

15.  x' -15x-\-50,x'-{-2x-S5,af-3x-  70. 

16.  a'-4:ab  +  4:b',a^-8  W,  a%  +  2  a^^^  4. 4  ab\ 

17.  m^  — 10  m^i  +  21  n^,  m^ —  5  mn  — 24: 71^,  m'^  —  Sln*, 

18.  a;2  +  5a;  +  6,  a^-2a;-8,  a^  +  2  a;2  +  5  a;  +  10. 

19.  9a6^-4a^6,  8  ac  +  2  ad-126c-3  6d. 

20.  a'-lQa,  a'-Sa^-4.a,  a' -{- 5  a^ -\- 4.  a. 

21.  27  n""  +  64  71,  18  n^  -  32  ^2,  9  n^  +  21 72*  + 12  711 

22.  9x2  +  300^  +  25,  6x''-{-7x-5,  10a:2-9a;  +  2. 

23.  71^-571  +  6,  9n^~n\  10-n-2n\ 

24.  aj3-7/3,  aj2_2a;2/  +  /,  x'-\-xy  +  y\ 

25.  3ac+ac«-6  6c-2  6rZ,  ac-4ad-2  6c  +  8  6fZ,  3c2-llcd-4c?2. 

26.  2x^-x-15,  2x^-7x-^3,  2aj2_9a,^9^ 

27.  a'-{-4.b'-9c^-4tab,a^-4:b^-9(^+12bc,a'-4:b'-{-9c^-6ac. 

28.  3  771^  +  m^Ti  —  2  mTi^,  6  771^71  + 11  mTi^  +  5  71^, 
9  m^n  +  5  771^71^  —  4  7W7i^. 

29.  32a«  +  4a^  12  a4  +  12  a^  +  3  a^  32a^  +  8a3  +  2a. 


FRACTIONS  103 


X.    FRACTIONS 

135.  The  quotient  of  a  divided  by  &  is  written  -  (§  6). 

h 

The  expression  -  is  called  a  Fraction ;  the  dividend  a  is  called 
h 
the  numerator,  and  the  divisor  h  the  denominator. 

l^he  numerator  and  denominator  are  called  the  terms  of  the 
fraction. 

136.  It  follows  from  §  69,  (3),  that 

If  the  terms  of  a  fraction  he  both  multiplied,  or  both  divided,  by 
the  same  expression,  the  value  of  the  fraction  is  not  changed. 


137. 

By 

the  Rule  of 

Signs 

in  Division 

(§  68), 

+  a 

+  & 

—  a 

-b 

-b 

—  a 

That  is,  if  the  signs  of  both  terms  of  a  fraction  be  changed,  the 
sign  before  the  fraction  is  not  changed ;  but  if  the  sign  of  either 
one  be  changed,  the  sigii  before  the  fraction  is  changed. 

If  either  term  is  a  polynomial,  care  must  be  taken,  on  chang- 
ing its  sign,  to  change  the  sign  of  each  of  its  terms. 

Thus,  the   fraction  ^~    ,  by  changing  the  signs  of  both 
c—d  b—a 

numerator  and  denominator,  can  be  written  — (§  51). 

d  —  c 

138.  It  follows  from  §§  124  and  137  that  if  either  term  of 
a  fraction  is  the  indicated  product  of  tivo  or  more  expressions, 
the  signs  of  any  even  number  of  them  may  be  changed  without 
changing  the  sign  before  the  fraction ;  but  if  the  signs  of  any  odd 
number  of  them  be  changed,  the  sign  before  the  fraction  is  changed. 

Thus,  the  fraction  ^^ may  be  written 

(c-d)(e-f) 

a  —  b  b  —  a  b  —  a,  .^ 


(d-c)(f-ey    (d-c)(e-fy        (d-c)(/-e)' 


104  ALGEBRA 


EXERCISE   48 

Write  each  of  the  following  in  three  other  ways  without 
changing  its  value : 

-     a       2   Vi±^.      3    _^ 4    2a;— 7^      ^         6x-5 


2  7  2-x  x-^2  {x-3)(y-\-4:) 

6.   Write    (^  ^  -  1)  («  -  ^)     i^    four    other   ways   without 
(x+5)(y-2) 

changing  its  value. 

REDUCTION  OF  FRACTIONS 

139.  Reduction  of  a  Fraction  to  its  Lowest  Terms. 

A  fraction  is  said  to  be  in  its  lowest  terms  when  its  numerator 
and  denominator  are  prime  to  each  other  (§  128). 

(We  consider  in  the  present  chapter  those  cases  only  in  which  the 
numerator  and  denominator  can  be  readily  factored  by  inspection. 

The  cases  in  which  the  numerator  and  denominator  cannot  be  readily 
factored  by  inspection  are  considered  in  §  444.) 

140.  By  §  136,  dividing  both  terms  of  a  fraction  by  the 
same  expression,  or  cancelling  common  factors  in  the  numera- 
tor and  denominator,  does  not  alter  the  value  of  the  fraction. 

We  then  have  the  following  rule : 

Resolve  both  numerator  and  denominator  into  their  fciCtorSf  and 
cancel  all  that  are  common  to  both. 

1.   Reduce     ^  ^  „  ^^o  to  its  lowest  terms. 
40  a^b^cH^ 

We  have. 


40  aWcH^     23  X  5  X  a^h'^cH^     5  cd^ 
by  cancelling  the  common  factor  2^  x  d'^lP'C. 

X?  —  21 

2.  Reduce  — to  its  lowest  terms. 

x?-2x-3 

By  §§  121  and  116,      x3  -  27     ^(x  -  3)(x^  4- 3^  +  9)^:.^  +  3x  +  9, 
^^^  'x2-2x-3  (x-3)(x  +  l)  «  +  l 

-    _,    .        ax  —  bx  —  ay-^by^    .,    ,         ,  , 

3.  Reduce to ? to  its  lowest  terms. 

b^  —  a^ 


By  §§  108  and  114, 


FRACTIONS  105 

ax-bx-  ay  +  by     (a  -b)(x  —  y) 


52  _  «2  (^)  4.  d)  (6  -  a) 

By  §  138,  the  signs  of  the  terms  of  the  factors  of  the  numerator  can  be 
changed  without  altering  the  value  of  the  fraction ;  and  in  this  way  the 
first  factor  of  the  numerator  becomes  the  same  as  the  second  factor  of  the 
denominator. 

ax-bx-  ay  +  by  _  (b  -  a)(y  -  x)  _ y  -x 

men,  52  _  ^2  -(& +  «)(&_«)-& +  «* 

If  all  the  factors  of  the  numerator  are  cancelled,  1  remains  to  form  a 
numerator ;  if  all  the  factors  of  the  denominator  are  cancelled,  it  is  a  case 
of  exact  division. 

EXERCISE   49 

Keduce  each  of  the  following  to  its  lowest  terms : 

.  5  xYz'        «  54  mri"           ^     126  a^6V  ^    90  aVn\ 

'  Sxfz^'         '  99  mV*            *      14  aV    *  *    36  amhi^' 

o  12  a'b'        -  63  a^j/V          g      26  m^ny  g     88  x'y'z^ 

'  42  62c3'         •  84  0.^2''           '    130m*nV*  '    66a^yz'' 

g  120  a^5VQ  ^Q    15x'y-\-10xY         ^  a;^-9a;  +  18. 

■  75a6V   '            ■     6  x'y^  +  4.  a^y' '            '  x''  +  x-12' 

12      a^  +  lla6  +  28  6^      ,  20      3a^-4a^-3a  +  4 

•  a«  +  14a'6  +  49a62*   '  '   9  a3  +  9  a^-ie  a-16* 

^3    64a^  +  72a;^?/4-8a;/  ^i    4m^  +  16myi  +  15n^ 

64  x^y  —  Sly^  '      6m^  —  mn  — 15  n^ 

14  ^^^  +  ^^^  ~  56  m^yt^     ,^  no    16  a;'^  +  4  a^  + 1 

w?  —  64  mn^  8  a;^  —  1 

15  a^  +  &'  23    a^-9/-2^  +  6y2! 
a2_2«^,_3ft2*  •   a;2_9^2^22_2a;2;' 

16  ac  +  3ad4-26c  +  66cZ    .  ^^    (a-2  5)^- (3  c -d/ 
3ac-ad+66c-26d*  '   (a  +  d)^- (2  &  +  3  c)^* 

.-        8a^-125  25    g^  +  28  ««&«  +  27  6« 

•  2a^  +  a;'-15a;*  *     a^ 4- 9  a^ft^ ^ 81  6*  * 

18  a^  +  a-12  /  og  25 -a^ 

•  3a2-13a  +  12'  '   ar^-lla;  +  30' 

19  (a^-49)(a^-16a;  +  63)         07       9a;^-49/ 
(«2-14a;+49)(a;2-2a;-63)*         '   28  a;/ - 12  a^2/ ' 


106 

28. 

Sb'-a^ 

a'-^-dab-Ub^ 

29 

27-a-          ^, 

4a;2_9a,_9 

ALGEBRA 


30.  tziSM^^, 

21 -a; -10x2 


31. 


15a;?/-20«-2l2/+28 


141.  Reduction  of  a  Fraction  to  an  Integral  or  Mixed  Expression. 
A  Mixed  Expression  is  a  polynomial  consisting  of  a  rational 

and  integral  expression  (§  63),  with  one  or  more  fractions. 

Thus,  a  +  -,  and  -  -\ ^  are  mixed  expressions. 

c  6       x  —  y 

142.  A  fraction  may  be  reduced  to  an  integral  or  mixed 
expression  by  the  operation  of  division,  if  the  degree  (§  64) 
of  the  numerator  is  equal  to,  or  greater  than,  that  of  the 
denominator. 

1.  Reduce  '^^ to  a  mixed  expression. 

ox 

By§72,      6x^4-15x-2^6x_^      15^_A^2x  +  5-A. 
•^  "*      '  3x  3a:       3a;      3x  3x 

2.  Reduce  — .  '    "^ to  a  mixed  expression. 

4a;2  +  3 

4x2  +  3)12x3-8x2  +  4a;-5(3x-2 
12  x^  +9x 

-  8  x2  -  5  X 
-8x2  -6 

-5x+l 

Since  the  dividend  is  equal  to  the  product  of  the  divisor  and  quotient, 
plus  the  remainder,  we  have 

12x8-8x2  +  4x-5  =  (4x2  +  3)(3x-2)  +  (-5x  +  l). 
Dividing  both  members  by  4x2  +  3,  we  have 

12x3  -  8x2 -4- 4x- 5 _  g ^ _ 2      -5x  +  l 

4x2  +  3  4x2  +  3  * 

Thus,  a  remainder  of  lower  degree  than  the  divisor  may  be  written  over 
the  divisor  in  the  form  of  a  fraction,  and  the  result  added  to  the  quotient. 

If  the  first  term  of  the  numerator  is  negative,  as  in  Ex.  2,  it  is  usual  to 
change  the  sign  o/  each  term  oj  the  numerator^  changing  the  sign  before 
the  fraction  (§137). 


FRACTIONS  107 


Thus,  12a^8_8:,2  +  4a;-5^3  _5x-l, 

'  4  x'^  +  3  4  x2  +  3 


EXERCISE  50 

Reduce  each  of  the  following  to  a  mixed  expression : 

1     15m2  +  12m-4                           o     30a«-5a^  +  15a-  +  7 
1. ^.   ^ 

3    9^+2  g        49a^  g    14a^  +  39a^  +  4a-19 

*3a;  — 1*  *7i»  +  32/*  *  2a  +  5 

'aj  +  2/  '    a  —  b  '  5x-\-2 

g     a^  +  8&^_  g    m^-nV         ^^    3a^  +  8a^-4    ^^ 

a  —  26  m  +  ri  a^-f-2a  —  3 

12    150)^-60.-^-20  a;^-7  ^3    24  a;^  +  21  a;  + 19 

3a;2-4  '  *      4o^-2aj  +  5 


14. 


6  g^  - 17  a'b  -  21  a'b^  + 19  a6^  +  22  b* 
2a^-5ab-6b^ 


143.  Reduction  of  Fractions  to  their  Lowest  Common  Denominator. 
To   reduce   fractions   to   their  Lowest  Common  Denominator 

(L.  C.  D.)  is  to  express  them  as  equivalent  fractions,  each  having 
for  a  denominator  the  L.  C.  M.  of  the  given  denominators. 

Let  it  be  required  to  reduce  — %-,  ;— ^^,  and  -—^  to  their 
T  ,  ^  •     4.  3  a^b^    2  ab^  4  a^b 

lowest  common  denominator. 

The  L.  C.  M.  of  3  a-b^,  2  ab%  and  4  a^b  is  12  a^b^  (§  133). 

By  §  136,  if  the  terms  of  a  fraction  be  both  multiplied  by  the 

same  expression,  the  value  of  the  fraction  is  not  changed. 

Multiplying  both  terms  of  ^^,  by  4  a,  both  terms  of  ^, 

by  6  a^b,  and  both  terms  of  — ^  by  3  b^,  we  have 

4a^6 

16  ace?    18  a^bm        -,  15  b^n 
12  a'b''  12  a'b''         12  a'b'' 


108  ALGEBRA 

It  will  be  seen  that  the  terms  of  each  fraction  are  multiplied 
by  an  expression,  which  is  obtained  by  dividing  the  L.  C.  D.  by 
the  denominator  of  this  fraction. 

Whence  the  following  rule. 

Find  the  L.  C.  M.  of  the  given  denominators. 
Multiply  both  terms  of  each  fraction  by  the  quotient  obtained 
by  dividing  the  L.  C.  D.  by  the  denominator  of  this  fraction. 

Before  applying  the  rule,  each  fraction  should  be  reduced  to 

its  lowest  terms. 

144.   Ex.  Reduce    ^  ^    and  — — — to  their  lowest  com- 

,  .     ,  a^  —  4         cr  —  oa-{-6 

mon  denominator. 

We  have,  a^ -4.  =  (a-{-2){a-2), 

and  a2-5a  +  6  =  (a-2)(a-3). 

Then,  the  L.  C.  D.  is  (a  +  2) (a -2)  (a  -  3).  (§  134) 

Dividing  the  L.  C.  D.  by  (a  +  2)  (a  —  2),  the  quotient  is  a  —  3 ;  dividing 
it  by  (a  —  2)  (a  —  3),  the  quotient  is  a  +  2. 
Then,  by  the  rule,  the  required  fractions  are 

, 4«Ca-3)  ^^^ Sa(a  +  2) 

(a  +  2)(a-2)(a-3)  (a +  2)(a-2)(a-3)' 

EXERCISE  51 
Reduce  the  following  to  their  lowest  common  denominator : 
..     Tab  Sbc  2ca  ^        4.a^  2 


6  '   10'    15'  ',  4a2-9'  6a^-9a 

5  4  6  a        1  8mn         2 


2«2 


m-n 


2       ^          ^          ^  6  

2m^n  5  7)i^n^'  7  mn^  m—7i  2{m—rif  Zim  —  iif 

o    3 3^4-42;  ^x  —  hy  m       3yi              5 

*    ~Wxf'     ^^yz"  '  '  71^-8' n2-4n  +  4* 

.     11  c'p   9a'm    Sbhi  g  2 3a 

■    12  a^ft' 14  6V  21  c^a"  *  a^ -{-3a' +  2a-{-6'  a' +  27' 

2  4          6 


x-\-2'  x-2'  x'-d' 


FRACTIONS  109 


^Q  a  +  36  a  — 3  b  a-\-4:b 


d'-7ab  +  12b''  a^-ab-Vlb'''  a"-9&2* 


a^4.3a;-10'  2a^  +  7a;-15'  2a^-7a;  +  6 


ADDITION  AND  SUBTRACTION  OF  FRACTIONS 

145.  By  §72,   ^  +  £-^  =  ^-±^1^. 

''  a      a     a  a 

We  then  have  the  following  rule : 

To  add  or  subtract  fractions,  reduce  them,  if  necessary,  to 
equivalent  fractions  having  the  lowest  common  denominator. 

Add  or  subtract  the  numerator  of  each  resulting  fraction, 
according  as  the  sign  before  the  fraction  is  -\-  or  —,  and  write 
the  result  over  the  lowest  common  denominator. 

The  final  result  should  be  reduced  to  its  lowest  terms. 

146.  Examples. 

.     o-      1-4^     4a  +  3  ,  1-662 

1.  Simplify      ■    ;,    +    ^    ,  3  * 

4a^6  Qab^ 

The  L.  C.  D.  is  12  a%^  ;  multiplying  the  terms  of  the  first  fraction  by 
3  62,  and  the  terms  of  the  second  by  2  a,  we  have 

4a  +  3      1-6&2  ^  12  a62  +  9  &^     2  g  -  12  ah"^ 
4a26  6a&3  12  a^ft^  12  a2&3 

^  12  a62  4.  9  52  _^  2  g  -  12  gftg  ^  9  62  +  2  q 
12g263  12g263 

If  a  fraction  whose  numerator  is  a  polynomial  is  preceded 
by  a  —  sign,  it  is  convenient  to  enclose  the  numerator  in 
parentheses  preceded  by  a  —  sign,  as  shown  in  the  last  term 
of  the  numerator  in  equation  (A),  of  Ex.  2. 

If  this  is  not  done,  care  must  be  taken  to  change  the  sign  of 
each  term  of  the  numerator  before  combining  it  with  the  other 
numerators. 

2.  Simplify  5^-4.v_7x-2y. 

^    •'         6  14 


110  ALGEBRA 


The  L.  C.  D.  is  42 ;  whence, 

6x-4y      7  x-2y  _36x-2Sy     21x-6y 
6  14  42  42 

_S5  X  -  28  y  -  (21  X  -  6  y) 

42 


(A) 


_36  X  -  2S  y  -  21  X  h  Q  y  _U  X  -  22  y  _7  X  -  11  y 
42  42         ~        21       ■ 


3.   Simplify 


oc^  -{-X      a?  —  X 


We  have,  x"^  ■\-  x  =  x{x  +  1),  and  x^^  —  x  =  x(x  —  1). 
Then,  the  L.  C.  D.  is  x(x  ■\-l){x-  1),  or  x{x'^  —  1). 
Multiplying  the  terras  of  the  first  fraction  by  x  —  1,  and  the  terms  of 
the  second  by  x  +  1,  we  have 

1 1      _     x-1 x  +  1 

x^-\-x     x^-x     x{x^  -  1)      x(a:2  -  1) 

__x-l-(x  +  l)  _x-l-x-l_      -2 

x(x^-l)  x(a;2-l)         x(x2-l)* 

By  changing  the  sign  of  the  numerator,  at  the  same  time  changing  the 

o 

sign  before  the  fraction  (§  137),  we  may  write  the  answer 

x(x2  —  1) 

Or,  by  changing  the  sign  of  the  numerator,  and  of  the  factor  x^  —  1  of 

2 
the  denominator  (§  138),  we  may  write  it 

4.   Simplify   __l__--_A_-_+  1 


a^-Sa  +  2     a2-4a  +  3     a^-Sa  +  G 

We  have,  a^  -  3  a  +  2  =(a  -  l)(a  -  2),  a^- 4a  +  3  =  (a- l)(a-3), 
and  a2- 5a  +  6=(a -2)(a- 3). 

Then,  the  L.  C.  D.  is  (a  -  1)  (a  -  2)  (a  -  3). 

Whence, \- 


a2_3aj_f.2      a2-4a  +  3     a^-ba  +  Q 

g-3 2(a-2)  a-1 

(a-l)(a-2)(a-3)      (a- l)(a  _2)(a  -  3)      («  _  i)(a-2)(a-3) 

a-3-2(a-2)+a-l_a-3-2a  +  44-a-l 
(a  -  l)(a  -  2)(a  -  3)         (a  -  l)(a  -  2)(a  -  3) 

«  -0. 


(a-l)(a-2)(a-3) 


FRACTIONS  111 


EXERCISE  52 

Simplify  the  following : 

1  4a;4-7      6x  —  5  a     2m-{-5n     3m  +  4yi^ 

10  15     *  '       8mV  6mn^ 

2  3  5  5     5a-7b     a+6b 
'    2a'b^     Ta'b'  '       27a  36b 

Q     4a  — 9     3a  — 8^  g     ^  — 2/  ■  y  —  ^^  i  g  — 3a; 

9  12  '      xy  2yz  3zx 

„     2(6n-{-5)      3(n  +  6)      4(5n-4) 

11  22  44        '  , 

g     3a-2     4a-7     7a-3 
3a«  7a2  9a    * 

g     8a;  +  l      10  y -9      9^  +  8 
7x  Uy  21  z   ' 

10  ^^'  +  3 _ 3  g-^  + 1  _ 3a^-2 

6  a'  12  a^  36  a* 

11  ^x-3     6x-{-5     5x-\-2     3  a; -10 

5  10  15  20      * 

io     3m-2     7m-8  ,  9m  +  4     lOm  +  7 
^^'    -4 6— +  -"8 9^' 

13     2a;4-?/     5a;  +  4y     8rg  —  3?/lla;  —  2y 
'    ~~8~~  16~"  24^  32 

14.  _^-+_-^.  18.    ^^ L. 

5m  — 2     2m  +  3  2x-\-y     2x  —  y 

15.  -i 1-.  19.        ^  ^ 


3a;-7     4a;  +  5  3a-9     5a-15 

16        m             2  20       5a;       4a;^  +  3a;  —  1 

m  +  2      m-2*  *    a;-3        a;2_^^_-L2 

y,     a  +  3      g- 3  2i     a;  +  3  j?/     a;  —  3  ?/ 

a  —  3     a +  3  x  —  3y     x-\-3y 

it 


112 


ALGEBRA 


25. 


33. 


22. 
23. 
24. 


a 

a 

a^  +  4  a  ■ 

-60     ar- 

-4a- 

-12 

X 

X 

2.^- 

-6 

2x-'d 

2x  +  3 

4.x'- 

-9 

a  —  n 

3a-4 

^^-l 

a-5 

71 

2a4-27i     3aH-3n     6a  +  6w 


2x 


2x 


Ux'-^x-^-l      16i»2-l 


26. 


3a;-f-2     9a^  +  4 
3x-2      9ar^-4' 


27. 
28. 
29. 
30. 
31. 
32. 


a^ 


+ 


a^  —  xy' 


x-\-y      (a;  + 1/)^      (x  +  yf 
a-\-h      .      0  —  6  2  ah 


2a-2h     2a  +  2b     a^'-h'' 


2a 


3a2 


a^ 


2a-l      (2  a- 1)2      (2a-iy 
m  — 1      m  +  l  .    m  — 6 


m-2 

X 


+ 


m+l      m ^ 

m-|-2      m^— 4 

a^  2a^ 


x-\-y     x^  —  xy-\-y^     a^-{-y^ 
b  .  c 


+ 


(a  +  6)(6  +  c)      (6  +  c)(c  +  a)      (c +  «)(«  + 6) 


+ 


2a 


a-3b     a-\-3b     (a-^Sbf 
1  .         1 


2  +  4a6  +  462     a2-462     a'-2ab 
5a;4-4     3a;-2     2a^-hl9a;-8 


a;2_3a;_4 


36. 
37. 

A    38 

2x2  +  5a;  +  3      4a;--f8a;H-3 


34. 


1 


27 


a;-f3     a^  +  27 


a;  — 4 
2 


+ 


a;  +  l 
3 


a-\-x     a—  X     a-\-2  X 
1  1 


FRACTIONS  113 

39        1  X  :ii?  -Q     g-l      g  +  l      a^~l 

■    x-l     iB2-l"^a^-l  '    a  +  l"^a-l     a^  +  l' 


41. 


4n-l  .         371+1 


6w2-17n4-12     lOw^-Qii-Q 


42      a-4       3a-l     5a^-9a4-ll 
•    2a-3       a  +  2         2a2  +  a-6 

43^        a?  +  4 a?-2         ^        a^4-3 


44. 


J^ 1 27yi 


1 1 2  mn  a-  /v*^ 


147.  In  certain  cases,  the  principles  of  §§  137  and  138 
enable  us  to  change  the  form  of  a  fraction  to  one  which  is 
more  convenient  for  the  purposes  of  addition  or  subtraction. 

1     o-       Tj;        3       ,  26  -fa 
1.    Simplify +       ^  ^' 

Changing  the  signs  of  the  terms  in  the  second  denominator,  at  the 
same  time  changing  the  sign  before  the  fraction  (§  137),  we  have 

3  26  +  <z 


a-h     a'^-b^ 

The  L.  C.  D.  is  now  a"^  -  b^. 

Then         — ^ 2fe  +  a^3(fl  +  6)-(2&  +  a) 

'        a-b     a^-b^  a"^  -  b^ 

__Sa  +  Sb-2b-a^2a  +  b 
a2  -  62  ^2  _  52' 

2.   Simplify  ^  ^ 


(x-y)(x-z)      (y-x)(y-z)      (z-x)(z-y) 

By  §  138,  we  change  the  sign  of  the  factor  y  —  x  in  the  second  denomi- 
nator, at  the  same  time  changing  the  sign  before  the  fraction  ;  and  we 
change  the  signs  of  both  factors  of  the  third  denominator. 

The  expression  then  becomes 

1.1  1 


(x-y)(x-z)      (x-y)(y-z)      {x-z){y-z) 


114  ALGEBRA 

The  L.  C.  D.  is  now  (x  —  y^ix  —  z)  {y  —  z);  then  the  result 

_(y-g)+(a;-g)-(a;-y)  _  y  -z  +  x-z-x  +  y 
(x-y)(x-z)(y-z)         (x  -  y){x  -  z){y  -  z) 

2y-2z  _  2(y-z)  5 


(x-y)ix-z)iy-z)      {x- y)(x  -  z)(j)  -  z)      (x-y){x~z) 

EXERCISE  53 
Simplify  the  following : 

4  1  g        a  a  2g? 


3a-3     2-2a 

2. 

3a;            2 
a;2_16     4-a; 

3. 

a-\-h           a  —  b 
a'-Sab     3b^-ab 

4. 

5           8m  +  6 
2m-l      l-4m2 

_     6x'-Sx-S2 

6. 


3  +  a3-a     a2_9 

4  3a;       2 

a^  —  x     1  —  x     X 


7.    _i__     1       I        ^-6 


71  +  4        1— ?2        71^4-371  —  4 

Q        3a  2a  8  a6 


a  +  26     26-a     a'-4:b^ 
2  1 


9a;^-16a;         4-3a;     x 


10.    , ^ -4- 


(x-y)(x-z)      {y-x){y-z) 


a3_63  a2  +  a64-62     &_« 

12.     „      3       .+   „     ^      .+     ^ 


a;^— 5a;4-6     a;^  — a;  — 2      4  — a;^ 
jQ     3m  +  l,    m  — 4        3m^  — 2m  — 4 


3m-l      5-2m     67?i2-17m  +  5 
14.  1  .  1  ,  1 


(a  —  b)(a  —  c)      (b  —  c)(b  —  a)      (c  —  a)(c  —  b) 


148.   Reduction  of  a  Mixed  Expression  to  a  Fraction. 
4  a;  —  t] 
a;  4-1 


-EJa;.     Eeduce  2  a;  —  3 — A-  to  a  fractional  form. 


FKACTIONS  115 

We  may  regard  2x  —  3  as  a  fraction  having  the  denominator  1,  and 
use  the  rule  of  §  145  ;  thus, 

2x      3     4x-5^C2a;-3)Ca;+l)-(4a;-5) 
x  +  1  X  +  1 

_  2x'^-a;-3-4a;  +  5  _  2x2 -5  a; +  2^ 
x+1  x+1 


EXERCISE  54 
Reduce  each  of  the  following  to  a  fractional  form : 

1.  ^  +  3  +  ^.  10.    2a.^-5a;  +  ME±iIl. 

4a;  4x4-9 

2.  2a-5-^^^.  11.   3a^-f-8-^^(^^-^). 

7a  ^  la-2 

3.  3n  +  4  +  --?_.  12.    ^^^±f-.(2x  +  y).L^ 

Sn  +  l  3a;-h52/ 

4.  ^-^^  +  1.  13.    -^4.-5 1. 


2n* 


a-f2m  a-|-6a  — 5 

K     -.      5a;  +  y  14.    m^-m^n  +  mn^-yiS- 

^2/  ^  I   97 

6.   5-3n  +  ^^i^.  ^a^4_3a;  +  9 

5  +  371 

7     2     2a-r6  16.   (^-2j£^  +  l. 

8.  ^  +  2.y  +  ,f  +  ^.      17.   M±^^+«^+2a+4. 

X  —  2y  ^  — ^a+-4: 

9.  a-46-^l±Mll  18..2a;+5y-    ^^  +  ^5.^^  . 

MULTIPLICATION  OF  FRACTIONS 
149.   Required  the  product  of  -  and  -. 

Let  |x|  =  ..  (1) 


116  ALGEBRA 


I 


Multiplying  both  members  by  b  x  d  (Ax.  7,  §  9), 


-X-Xbxd  =  xxbxd,    or 
0      d 


ix')%>^'^y->^^xd 


for  the  factors  of  a  product  may  be  written  in  any  order. 

Now  since  the  product  of  the  quotient  and  the  divisor  gives 
the  dividend  (§  67),  we  have 

-  xb  =  a,  and  -  x  d  =  c. 
b  d 

Whence,  (a)  x  (c)  =  x  xb  x  d. 

Dividing  both  members  by  6  x  c?  (Ax.  8,  §  9), 

b  xd 
From  (1)  and  (2),  ^  x  ^  =  ^.  (Ax.  4,  §  9) 

Then,  to  multiply  fractions,  multiply  the  numerators  together 
for  the  numerator  of  the  product,  and  the  denominators  for  its 
denominator. 

150.  Since  c  may  be  regarded  as  a  fraction  having  the 
denominator  1,  we  have,  by  §  149, 

-XC--X--- 

Dividing  both  numerator  and  denominator  by  c  (§  136), 

a ,,  a 

X  c  = 


b  b^c 

Then,  to  multiply  a  fraction  by  a  rational  and  integral  expres- 
sion, if  2^ossible,  divide  the  denominator  of  the  fraction  by  the 
expression;  otherwise,  multiply  the  numerator  by  the  expression. 

151.  Common  factors  in  the  numerators  and  denominators 
should  be  cancelled  before  performing  the  multiplication. 

Mixed  expressions  should  be  expressed  in  a  fractional  form 
(§  148)  before  applying  the  rules. 


FRACTIONS  117 


1.   Multiply  -^  by  ^^. 


10  a^y  ^  3  6^g3  _  2  X  5  X  3  X  a^b^x^y  _  5  6%  _ 
9  6a;2      4  a^y^       3^  x  22  x  a^ftxV         6  y  ' 

The  factors  cancelled  are  2,  3,  a^,  &,  x^,  and  y. 

2.   Multiply  together -^j±2^,  2-^-^^,  aud  J^f. 

_5l±l£_x  /o      a;  -  4\      x2  -  9 


x2  +  2  X    ^  2a;-6-a;  +  4  ^  a;^  -  9 


X2  +  X-6  x-3 

^       x(x  +  2)        ^  a;  -  2  ^  (X  4-  3)  (x  -  3)  ^     x 

(x  +  3)(x-2)      x-3      (x  +  2)(x-2)      x-2* 

The  factors  cancelled  are  x  +  2,  x  —  2,  x  +  3,  and  x  —  3, 

3.  Multiply  ^±^^  hj  a-b. 

Dividing  the  denominator  by  a  —  6,  ^-^^ —  x  (a  —  6)  =  ^— i 

a^  —  62  a  +  6 

4.  Multiply  —HL-.  by  m  +  w. 

m  —  n 

Multiplying  the  numerator  by  m  +  n,      ^     x  (m  +  n)  =  «*!_+:^. 


EXERCISE  55 

Simplify  the  following : 

*.    8cd«       35a'b''  '    15a«      126^      7c** 

«     5  a^      9  6^  ^  7  c*  c      28  m^        15  n«  5  a^ 


3  6^     10  c^     6  a*  25  nV     14  mV     21  mV 


118  ALGEBRA 


7.    ^^x(2a-6).  9.    ^i!fl|?x        '-' 


a-2b  '  47i2        n'  +  7i-^2 

8.    _^-Zl2         ^3^^  ^^j    a2_2a-35     4a3_9a 

12     ^«^-5^x     «'  +  3«  +  ^ 


14. 


16. 


18. 


a^-27       4a2-20a  +  25 


4m^  +  8m  +  3     6m^-9m 
2m2-5m  +  3       4m2-l 

25     «^  -\-  mx  4-  na;  4-  mn     a^  —  w? 
x^  —  mx  —  nx-{-  mn      ^  —  v? 

a''-2ab  +  b^-c^     a-^b-c 
a'-\-2ab-\-b^-c^     a-b  +  c' 

17     16^-4^20^+5^a;2  +  2a;  +  l 


5x-{-5       6x-^6         16ar^-l 
a2-lla4-30^  a2-3a^,        a' -9 


\  2a+3        7^  4a-5       y 

a^-Sf     x  +  2y      \   ^x'-2xy  +  4:yy 

^21     9a;^  +  12aa;  +  4a^       a;'^  +  aa^  ^/        2a^  +  2aa;-a^ 
x'-a'  3a;  +  2a      V  3x  +  2a      J 

22      2n^-n-S       n^4-4n  +  4     n^-yi-2 
n^_8n2  +  16  n^  +  n  2n2_3n' 

DIVISION  OF  FRACTIONS 

152.  Required  the  quotient  of  ^  divided  by  ^« 

6  a 


FKACTIONS  119 


Let  1^1  =  ..  (1) 

Then  since  the  dividend  is  the  product  of  the  divisor  and 
quotient  (§  67),  we  have 

a     c 


h     d 

Multiplying  both  members  by  -  (A x.  7,  §  9), 

c 

bed            c 

(2) 

From  (1)  and  (2),         ^^l  =  ^x^- 
babe 

(Ax.  4,  §  9) 

Then,  to  divide  one  fraction  by  another^  multiply  the  dividend 
by  the  divisor  inverted. 

153.  Since  c  may  be   regarded  as  a  fraction  having  the 
denominator  1,  we  have,  by  §  152, 

a  a     1      a 

-^c  =  -x-  =  — • 
b  b      c      be 

Dividing  both  numerator  and  denominator  by  c  (§  136), 

a  a-T-c 

Therefore,  to  divide  a  fraction  by  a  rational  and  integral 
expression : 

If  possible,  divide  the  numerator  of  the  fraction  by  the  expres- 
sion ;  otherwise y  multiply  the  denominator  by  the  expression. 

154.  Mixed  expressions  should  be  expressed  in  a  fractional 
form  (§  148)  before  applying  the  rules. 

1.   Divide  1^  by  li^. 

Wehave         6q2fe       9^253  ^  6^25  ^^  lOa;^^  4y8 
*         SxV  •  iOa;V      5a.3y4      9^253      352^; 


120                                         ALGEBRA 
2.   Divide  2 —  by  3 — -— 


_2x  +  2-2a;  +  3  .  Sx^  -  3  -  3a;2  +  13 
X  +  1  ■  x2  -  1 


r2 


l_5(x  +  l)(x-l)_x-l 


x  +  1         10         2x5x(x+l)         2 


3.  Divide  ^?L^bym-n. 


Dividing  the  numerator  by  m  -  w,  ^^  ~  ^^  --  (m  -  w)  =  ^'^  "^.^^  t  ^'^- 

4.  Divide  ^^  bya  +  &. 

a  —  b 

/j2  4-^2  <22    I    7)2 

Multiplying  the  denominator  by  a  +  6,  — — ^  (a  +  &)  =  — ^— -• 

If  the  numerator  and  denominator  of  the  divisor  are  exactly 
contained  in  the  numerator  and  denominator,  respectively,  of 
the  dividend,  it  follows  from  §  149  that  the  numerator  of  the 
quotient  may  he  obtained  by  dividing  the  numerator  of  the  divi- 
dend by  the  numerator  of  the  divisor ;  and  the  denominator  of 
the  quotient  by  dividing  the  denominator  of  the  dividend  by  the 
denominator  of  the  divisor. 

5.  Divide  ^^-iy\y3x  +  2y^ 

ar  —  y^  *  x  —  y 

We  have,  9x2 -4y2^3x  +  2y  _3x  -  2y. 

x2  —  2/2  x  —  y         x  +  y 


EXERCISE  56 

Simplify  the  following: 

.     45  a: V  .  o   ,    ,  '    2    12  a%'  .   9a^&« 

*     4tfn    '  '  '   55<fd'  '  22  c'd' 


FRACTIONS  121 

'      a-3        ^  ^  a2  +  6a  +  9        aH-3 

4.   ^  —  ^y^(x  +  3y).  9.  f^  +  ^^f^-^. 

g     n2-3n-40  ,  yi2  +  4n-5  ^^  a^-3a;y  .  a;^-10a:y+21y^ 

4:%           '          5n  '  a^  —  y^   '     x^-{-xy-\-y^ 

'       a»-9a62      '  a  +  36*  '          4a;-3         •  <.   ^+   > 

7       9a^-4?/^     .    9a^  +  6ir.y  j^g  8yi«  +  l    .  4n^-2n  +  l 

16a^-25/  '  Sxy-lOy^'  '  2n^-^4.n  '    n2  +  4n4-4  ■ 


13.      2 


2  +  8a;-3a;' 


9 


H'-W)- 


14.    (x'-y2  +  2yz-z')^^~y~^^. 

x-\-y  +  z 

j5     m^  +  2m^  +  m  +  2  .  m^  +  3m^H-2 
m^  — m^  — m  +  l    *  m"*  — 2m^  +  l 

jg    2a^-a6-36^  .      3a^  +  a6-2&^ 

9a2-2562    •  9a2-30a6  +  2562' 


COMPLEX  FRACTIONS' 

155.  A  Complex  Fraction  is  a  fraction  having  one  or  more 
fractions  in  either  or  both  of  its  terms. 

It  is  simply  a  case  in  division  of  fractions;  its  numerator 
being  the  dividend,  and  its  denominator  the  divisor. 
a 


1.    Simplify 


=  axo^(§152)=    '^ 


h—-     ^^  —  ^  bd  —  c^         ^     bd 


It  is  often  advantageous  to  simplify  a  complex  fraction  by 
multiplying  its  numerator  and  denominator  by  the  L.  C.  M.  of 
their  denominators  (§  136). 


122  ALGEBRA 

a a__ 

2.'  Simplify  — —' 

a—b     a+b 

The  L.  C.  M.  of  a  +  &  and  a-b  is  (a  +  b)(a  -  &). 
Multiplying  both  terms  by  (a  +  6)(a  —  6),  we  have 

a a 

a-h     a  +  b  _  a(a  +  6)-  g  (g  -  &)  _  a^ -\-  ab  -  a^  -h  ab  _    2  ab 

b     j^     a     ~h{a  +  b)+a{a-b)      ab  +  b'^  +  a^  -  ah     a^  +  b'^' 
a  —  b     a-{-b 

3.   Simplify ^ — 

1+     ^ 


1+i 

X 


a;  +  l  05  +  1 


-.  1         ■■    ■      X         x  +  l-\-x     2x+l 

1+1      ^+1 

In  examples  like  the  above,  it  is  best  to  begin  by  simplifying  the  lowest 
complex  fraction. 

1  X 

Thus,  we  first  multiply  both  terms  of by  x,  giving  — — - ;  and 

111  X  -r  1 


1  ^  x  +  1 

then  multiply  both  terms  of —  by  x  +  1,  giving  • 

-i.X  X+i+X 

x  +  1 

EXERCISE  57 

Simplify  the  following : 

X     y  9a  m 

x  —  y_x-\-y  a-\-b  _     5 

a;4-y      a;  — y  '  ?4-?: 

X  y  a      b 


FRACTIONS 


123 


6. 


3a    .    fy    .   4rb 


a     0 


21j 

X 


^4.1_^ 


x^y^ 


+ 


xy       ^1-xy 


9. 


10. 


l-\-xy  xy 

^ 1_ 

1  —  a;      1  -\-x 


1-a^     1+a^ 


14. 


15. 


16. 


17. 


18. 


27  g^     h 
6^       a 

h  a 

4a^  — 2/^ 
3      x-4.y" 
2x-y 


a;  +  4 


ic  — 4 


5ic—  1      5a;  +  1 
a; 4-4        ic  — 4  ' 


5a;-l 


2yz 


5a;  +  l 
L^-1 


a:^  +  y^ 


2xy 
3a 


(a  +  2)^     a  +  2 
2  a^  +  2  a  - 1         c 


( 


a-2 


11.  1- 


1+a 


13. 


^ 


^a 


4a-l 


1- 


2a  +  5 
3a-2 


3ar^H-y^     a;  +  y 


/ 


124  ALGEBRA 

a  —  oi  a?  —  Q? 

2j-    a-{-x  g^  +  a^ 

a  —  x  g^  — a^' 

a  +  a;  a^  H-  a^ 

MISCELLANEOUS  AND  REVIEW  EXAMPLES 
EXERCISE  57  a 

Reduce  each  of  the  following  to  a  mixed  expression : 

J     ^x'-2a?-20        2     m'  +  n\       ^    12  g^-3  g^-22  g  + 8 
12  a^         *         '    m  —  n  '  3a^  —  5 

Simplify  the  following : 

^•5^:x^^-^^>     ^^•5^-1^.^67- 

^-    ;:rTT4"^('^  +  ^)-  a;*-2a^  +  8a;-16 


g*4-w* 


2  ma;  /n  o  „\  13. 


(a_6)2_(c-d)^ 


7.  ^  ""^"     ^(2m-3a;).  (6  +  d)2 _  (a  +  cV 

_1_,1_       _! L 

8.  a^-5a;-84  .    a;4-7  ,  14.     2a;"^3y       3a;     2y 

27a;»-8        3a;-2  4.x^-9y^'^  9  x'-4.y^' 

9    3a;  I  1     ^^  +  ^-  15    g^6cZ  -  g6^c  -  acd^  4-  &c^^. 

2  a; -3  '    hhd  +  ahc"  -  abd^  -  d'cd 

16     /«  +  2         2    Y    g 3\ 

■    (^     g    '^g_3Aa-2     a  +  s) 


17. 


fl^_2ar^4-2a;-l 
a;^  +  ar^+l 


jg  _J_  ,_i § llg-56. 

'  a  —  h      b  —  a     g+ft        W—a^ 

19  27  a;^  4-1  .      15a;^-a;-2 

■  25a;2-4  *  25 a;^ - 20 a;  +  4* 


FRACTIONS  125 


20     (2      ^  +  4a^-2n  .  (x  +  1      x-^\ 
'    [^         a^^2x-Sj  '  [x-2^x  +  4.J 

ft,  2c~d  c-\-2d 


ac-^2ad-^2bc  +  4:bd     2  ac-ad+4.bc -2bd 

\^                     a;H-4  J      \                x-\-4:  J 
23  0^  +  3  ,       a  +  2 g  +  l 

25.  t:^ ?^=:^ 

x^-~{y-zy      (x-zf-y^ 

og       (a+26  +  3cy-(2a-3&-cy 
•    (3a4-6-2c)2-(2a  +  66-h2c/ 

27.       1-1-^+     ^ 


X  +  3     X  —  3     x-\-4      x  —  4: 

(First  combine  the  first  two  fractions,  then  the  last  two,  and  then  add 
these  results.) 

28  3  3  5n'  5  n' 

271  +  1      271-1      8n«  +  l      Sn^-1 

30  &  -  c  c  —  g a—b 

'    (a-b)(a-c)      (b-c){b-a)      {c  -  d)(c  -  b) 

3.     (2x^-\-5x-2y-25^ 
'    (3a^-4(»-3)2-16* 

32  ft      I      1  3         a^  +  2a 

•    a  +  3      a-3      a2-9       a^  +  O  ' 

(First  add  the  first  two  fractions,  to  the  result  add  the  third  fraction, 
and  to  this  result  add  the  last  fraction.) 

33  3q         3a  C^a'     ,    12  a* 


a+b     a-b     a'^-y'     a* +6* 


(§ 


126  ALGEBRA 


'^      >34.    --A.-----J—^  + 


a -2      2  a^  +  4 


2(a-l)      2(a  +  l)      a^  +  l        o>-\ 
35  1 1  ,  1 


39. 


_2 1       1  2 


2  a;  +  2/  3^  +  2?/ 


3g        6  g^  -  g  -  2         8a^-18a-5  4a^-9 

4a2-16a+15     12a2-5a-2     4a2  +  8a  +  3' 


(SiJ- 


a;  +  1\'     /a?  - 1 


i) 


.^_2      a;-3     ic-4     .  (a;  -  2)  (aj  -  3)  (a;  -  4) 


40     a^  +  2      a;  +  3     a;4-4     3  a^- 10  a^- 21  a;  + 66 


FRACTIONAL  AND  LITERAL  LINEAR  EQUATIONS    127 


XI.     FRACTIONAL     AND     LITERAL    LINEAR 
EQUATIONS 

SOLUTION  OF  FRACTIONAL  LINEAR  EQUATIONS 

156.  If  a  fraction  whose  numerator  is  a  polynomial  is  |)re- 
ceded  by  a  —  sign,  it  is  convenient,  on  clearing  of  fractions, 
to  enclose  the  numerator  in  parentheses,  as  shown  in  Ex.  1. 

If  this  is  not  done,  care  must  be  taken  to  change  the  sign  of 
each  term  of  the  numerator  when  the  denominator  is  removed. 

1.  Solve  the  equation —  =  4  H --^ — 

4  5  10 

The  L.  C.  M.  of  4,  5,  and  10  is  20. 

Multiplying  each  term  by  20,  we  have 

15a;- 5-  (16X-20)  =80  +  14x  +  10. 
Whence,  15  a;  -  5  -  16  x  +  20  =  80  +  14  a:  +  10. 

Transposing,  15  a:  - 16  x  -  14  x  =  80  +  10  +  5  -  20. 

Uniting  terms,      »  —  15  x  =  75. 

Dividing  by  —  15,  x  =  —  5. 

2  5  2 

2.  Solve  the  equation =  0. 

^  x-2     x-\-2     af-4: 

The  L.  C.  M.  of  X  -  2,  X  +  2,  and  x^  -  4  is  x2  -  4. 
Multiplying  each  term  by  x^  —  4,  we  have 

2  (X  +  2)  -  5  (x  -  2)  -  2  =  0. 

Or,  2  X  +  4  -  5  X  +  10  -  2  =  0. 

Transposing,  and  uniting  terms,  —Sx=  —  12,  and  x  =  4. 

If  the  denominators  are  partly  monomial  and  partly  poly- 
nomial, it  is  often  advantageous  to  clear  of  fractions  at  first 
partially  ;  multiplying  each  term  of  the  equation  by  the  L.  C.  M. 
of  the  monomial  denominators. 


128  ALGEBRA 

o    a  1      ^v  ^-       6x-\-l      2x-4:      2a;-l 

3.    Solve  the  equation  — -^ -—  =  — ;: — • 

^  15         7  a; -16  5 

Multiplying  each  term  by  15,  the  L.  C.  M.  of  15  and  5, 
a      ,   ,      30  5C-60      ^„      o 


^"'^       lx-16-^'^      "• 

Tra-nsposing,  and  uniting  terms,  4  =  '     ^~..^  • 

7  a:  —  16 

Clearing  of  fractions,         28  ic  -  64  =  30  x  -  60. 

Then,                                         —  2  a;  =  4,  and  x  = 

-2. 

EXERCISE  58 

In  Exs.  5,  11,  22,  and  32,  of  the  following  set,  other  letters 
than  X  are  used  to  represent  unknown  numbers. 

This  is  done  repeatedly  in  the  later  portions  of  the  work. 
Solve  the  following  equations : 

.      1__1  =  ^ ]_  2     -i ? ?-.=,-L. 

'    2     9x     9     6x  '    5x     10a;     15a;         12* 

3     A_ J__   ^         7    ^11 

'    3  a;     12  a;     8  ac     24  x  8  * 

4.  4^  +  ^^±l  =  -^.  7.   a;-i^  +  ^^±^  =  -2. 

5  2  -38 

5.  ^  +  3_?rt^  =  2v.  8.    3a^  +  2     5a;-6^5 

5  15  *  *        5  8  4 

6     Z£_§_£ZL?_2  =  —  9     ^^-1^     7a;+4    3a;-8__Q 

■     4  7  14*  *         9  12  8  ' 

.Q    5(x-l)     2(a;  +  2)^,      5a;-15 
6  3  4 

„     lli)  +  12     4p-6     5p-9^      o        , 
18  9      "^      4 

12     8a;-l      ll'a;-7      13  a;  +  3^14  a;  +  38 

3  5  10  15      *      . 


FRACTIONAL  AND  LITERAL  LINEAR  EQUATIONS    129 

13  5a;  +  4      16  a;-h5^  10^^-9      4(3a;-2) 

3  9  5  15       ' 

14  3(a;  +  7)      7x  +  10^4:X-7     2(7x-l) 

7x  Sx  6  21        ' 

..     (3a;-4)(3a;  +  l)      (8  a;- 11)  (a;  +  l)^  (5  a;- 1)(4  a; -3) 
2  4  8  * 


16. 


A_  =  o..         18.   21^!±I^±11=.3. 


4a;-3      7a;-3  7iB2_4a;-9 


17     6a;  +  1^2a;  +  3  ^^    2a;  +  7_   10a;-3 


22. 


9x-5     Zx-2  a;2_4      5a;(a;  +  2) 

OQ     8a;  +  57^2a;-15      2a;  +  16 
12  aj  +  8  3 

o.     12  a;  — 5        3a;  +  4    ^4a;  — 5 
21  3(3a;-fl)  7 

3/1-1      5y?.  +  4_      o      2^    ?^JIl?  =  a;_l_JL^±iL. 
n-5         71  +  8  *         '9  3(3a;  +  4)* 

23     6a;^  +  23        a;-l   ^^      05    5a;-4     2a;  -  7^4ar^- a; 

'    (2a;-3)2     2a;-3        *  *       7  14  7a;-2* 

26.   ? i 4- ^- =  0. 

2a;  +  l     3a;  +  2     6ar^  +  7a;  +  2 

07  1 ^  I  1  _Q 

6a;-24     6a;  +  3     6a;-4       * 


28. 


4a;^  +  3a;  +  2^6a;^-5a;-4 
4a;  +  3  6a;-5 


29  8a;^  +  4   _     hx  3a; 
16a;2_25-5  +  4a."^5-4a;* 

30  7a;  5a;   ^  12(a;^-l) 
a;  +  3      1-a;     a;2_^2a;-3* 


31. 


2ar^-a;  +  3     2a;^-f-3a;-l^-20a;^-6a;4-3 
3a;  +  2  3a;-2  9ar^-4 


130  ALGEBRA 


22     3  a:  — 5  _  4a;  +  2  _  15  a;  — 1      7 

2  3a;  +  2~      10  5* 


34. 


x  +  2     x  +  3~x-i-6     x  +  T 


(Eirst  add  the  fractions  in  the  first  member ;  then  the  fractions  in  the 
second  member.) 

!         x-\-l      x  —  6     x  —  7 


35. 
36. 


x  —  2     x  —  1     x  —  4:     x  —  5 

4aj  +  7      8«  +  4      12£c  +  l        5a;-l 


5  15  45  9(5a;4-2) 


37       a^'  +  S  1       _        2«-l 


2(a;3_8)      6(03-2)      3(a;2  +  2a;  +  4) 
«Q     2a;-l      1  ,    2a;  +  3    ,  5«2_|.30^ 


a;-2       2     3a;  +  10     2 (a; -  2) (3 a;  + 10) 

39     2a;  +  l     5a;-6^2  23a;^-10 

'    3a;-5     2a;  +  7  635^  + 11a; - 35* 

157.   Solution  of  Special  Forms  of  Fractional  Equations. 

1.  Solve  the  equation  |^^  +  K^  =  2. 

^  2a;-3      a;2_^4 

We  divide  each  numerator  by  its  corresponding  denominator ;  then 

14-— ?— +  1-^^+1  =  2    or-A__^±i  =  o 
2x-3  a;2  +  4       '        2x-3     «2  +  4 

Clearing  of  fractions,  2  a;2  +  8  -  (2  a:2  4.  5  a;  -  12)  =  0. 

Then,  2a:2  +  8-2a;2-5x  +  12  =  0;  whence,  a;  =  4. 

We  reject  a  solution  which  does  not  satisfy  the  given  equation. 

2.  Solve  the  equation +  "" 


x-S       x-2       0.-2-535  +  6 
Multiplying  both  members  by  (x  -  3)(x  —  2),  or  aj2  —  5  as  +  6, 
x-2  +  «-3  =  3x-7. 


FRACTIONAL  AND  LITERAL  LINEAR  EQUATIONS    131 

Transposing,  and  uniting  terms,  —x  =  —  2,  or  x  =  2. 

If  we  substitute  2  for  x,  the  fraction  becomes  — 

x-2  0 

Since  division  by  0  is  impossible,  the  solution  x  =  2  does  not  satisfy 

the  given  equation,  and  we  reject  it ;  the  equation  has  no  solution. 

3             4            2^' 
"3.   Solve  the  equation  —  -\ = 4- 


«  +  10     »  +  6     aj  +  8     x  +  9 

Adding  the  fractions  in  each  member,  we  have 

7x  +  58        _       7a;  +  58 
{x  +  10) (X  +  6)  ~  (X  +  8) (X  +  9)* 

Clearing  of  fractious,  and  transposing  all  terms  to  the  first  member, 

(7x  +  58)(xH-8)(a;  +  9)-(7x  +  58)(«  +  10)(x  +  6)  =  0.  (1) 

Factoring,       (7  a;  +  58)  [  (x  +  8)  (x  +  9)  -  (x  +  10)  (x  +  6)  ]  =  0. 
Expanding,  (7 x  +  58) (x^ +  17 x-\-12 -x^ -Wx- 60)  =  0. 

Or,  (7x  +  58)(x4-12)  =  0. 

This  equation  may  be  solved  by  the  method  of  §  125. 
Placing  7  X  4-  58  =  0,  we  have  x  =  -  ^•• 
Placing     X  + 12  =  0,  we  have  x  =  — 12. 

158.   If  we  should  solve  equation  (1),  in  Ex.  3  of  §  157,  by- 
dividing  both  members  by  7  a;  +  58,  we  should  have 

-       (x-{-S)(x  +  9)-(x  +  10){x-{-6)  =  0. 

Then,       a^  +  17  a;  +  72-a^-16  aj-60  =  0,  or  cc  =  -12. 

In  this  way,  the  solution  ic  =  —  -^  is  lost. 

r  It  follows  from  this  that  it  is  never  allowable  to  divide  both 
I  members  of  an  equation  by  any  expressioii  which  involves  the 

unknown  numbers,  unless  the  expression  be  placed  equal  to  0  and 
\the  root  preserved,  for  in  this  way  solutions  are  lost, 

EXERCISE  59 

Solve  the  following  equations : 


J    2£-f7     2^-^^g  2       4a;  4- 11 


2aj-Hl      x-2  x'-{-x-20     x  +  5     a;~4 


132  ALGEBRA 

3  _8 3_^J^ 5_       g    2x-j-S  2a;-3        36     ^^ 

a;+3     x-7     x-\-9     x-2         '   2x-d  2x^-'^    ^:^-9~ 

4  a;  +  3     a;  +  4     a^  +  2^g        .^    2a;+5  3a?^H-24a;4-19^     .^^ 

a;  +  2     a;  +  3     a;H-4        *         *     a;+7  a;2+8a;+7 

r       3.2  1.4 


,      4         e    a;2_2a.4.5     a^_^3a._7 


a;+9     a;+4     a;+3     aj+18  0.-2—20;— 3     x2+3a;+l 


SOLUTION  OF  LITERAL  LINEAR  EQUATIONS 

159.   A  Literal  Equation  is  one  in  which  some  or  all  of  the 
known  numbers  are  represented  by  letters ;  as, 

2a;  +  a  =  62_^10. 

Ex.  Solve  the  equation  — ^^^  =  ^^±1'. 

x  —  a      x-\-a      01?  —  (T 

Multiplying  each  term  by  x^  —  a^, 

x{x  +  a)  -  (x  +  2  6)  (x  -  a)  =  a2  +  62, 
or,  x2  +  ax  -  (x2  -I-  2  &x  -  ax  -  2  a&)  =  a^  +  62, 

or,  x2  4-  ax  -  x2  -  2  6x  +  ax  +  2  a6  =  a2  +  6^, 

or,  2  ax  -  2  6x  =  a2  -  2  a6  +  62. 

Factoring  both  members,  2  x(a  —  6)  =  (a  —  6)2. 

Dividing  by  2(a  -  6),  x  =  |^^=^  =  ^• 

In  solving  fractional  literal  equations,  we  must  reject  any  solution 
which  does  not  satisfy  the  given  equation.     Compare  Ex.  2,  §  157. 

EXERCISE  60 

Solve  the  following  equations : 

1.  (aiB-6)(6a;  +  a)  =  6(a«2-6). 

2.  (a;-2a-6)2=(a;  +  a  +  26)2. 

3        3a;         x  —  2n_fy  a     4a;H-3a      6ct  +  56_q 

2x  +  n         2x  '  '    4a;— 3a     6a  — 5b 


FRACTIONAL  AND  LITERAL  LINEAR  EQUATIONS    133 


.    — +  — +  "-  =  a4-&  +  c. 
ao      00     ca 

6     a7(a  +  4  6)  — 6^     x—b  _x-j-a 
a^  —  b^  a-\-b     a  —  b 

y     m^x-\-n  _  n^x  +  m  _m  —  n 
mx  nx  mnx 

g     X  —  a     x  —  b     X  —  c _bG(x -\-  b)  —  ah^  —  a^c -\- abx 
b  c  a  abc 

Q  5 2 3  m 

2a)  +  5m     3 a; - 4  m ~ 6 a^  +  7 ma? - 20 m^* 

10  ^  +  6  I  g  — 26_(2a  — 6)a;  +  3a6 
'       a;  a;  +  a  oc^  —  a^ 

11  ^^     g^  +  ?>^  _  g^     a;  (g  —  6) 


12. 


g  g2         6^  & 

&         g  —  6 


a;  —  g     x  —  b     X 


13_  «(»-«)+H^-&)^„  +  6. 

X  —  b  x  —  a 

14.  (g  +  6)(a;-g  +  6)-(a-6)a;  +  g2-62  =  2g(a;  +  g 

15.  (x -j-p  +  g)(a;  -p  +  g)  +  ^2  ^  (x  -p)(x  +  g). 

16     ^^  +  ^^     4:X  —  5n_        10  n^ 

x-\-2n        Sn  —  x      a^  —  nx  —  6n^ 

jiy     3a;     5ga;  —  26^a  +  3  6a;     ax-{-2a? -~^h^ 
'2  4  a  8?)  1fia/>         * 


6). 


16  a6 

b  a—  b 


jg     _a o___     g— 0 

a;4-6     x-[-a     a;H-g  +  6 

iQ      a;  +  g        a;  —  g         2ga;  —  19g2_^ 
«  —  2g     a;-|-3g     x^-\-ax  —  Qa? 

20.  -J^ i_=_I L_. 

a;— 2a     6a;  +  g     3a;  — 8g     2x  —  Sa 


134  ALGEBRA 


21.        4  14  1 


x  —  4:n      op-\-n      x-\-4:n      x-\-3n 

22      a^  —  2  aa?  +  ft^    ,  x^-\-ax  —  2a^  _  q 
x^  —  2ax  —  Sa^     x^ -\- ax -\- 2  a^ 

no     x-\-a     x  +  b     x  —  a  —  h _  o 
x  —  a     x  —  b     x-{-a-\-b 

24.   x''-^(x-ay+(x-by  =  3x(x-a)(x-b). 

SOLUTION  OF  EQUATIONS  INVOLVING  DECIMALS 
160.   Ex.     Solve  the  equation 

.2  a?  +  .001  -  .03  a;  =  .113  a^  -  .0161. 

Transposing,      .2x-  .OZx-  .113  x  =  -  .0161  -  .001. 
Uniting  terms,  .057  x  =  —  .0171. 

Dividing  by  .057,  x  =  -  .3. 

EXERCISE  61 

Solve  the  following  equations  : 

1.  7.98 a?- 3.75  =  .23 a; +  .125. 

2.  3  a; +  . 052  -  7.8  a?  =  .04- 5.82  a;-. 0696. 

3.  .05i;-1.82-.7'y  =  .008v-.504. 
(Here,  v  represents  the  unknown  number.) 

4.  .73 x  + 8.86  =  .6(2.3 a; -.4). 

5.  .07(8aj-5.7)  =  .8(5a:  +  .86)  +  1.321. 

6.  3.2 X-. 84  +  -^^ ^--^^^  =  .9 a;. 

.9 

„     6.15  a; +  .67     .6aj-.81     5 


3a; 

—  ( 

.3  ~J 

-3.1 

.6  .03 


V3     .9a;-2.84     .8a;-6.52^^^ 


9.   20.1a.-:ni^=lM§?^^:^_.135. 


FRACTIONAL  AND  LITERAL  LINEAR   EQUATIONS    135 

PROBLEMS  INVOLVING  LINEAR  EQUATIONS 

161.  The  following  problems  lead  both  to  integral  and  frac- 
tional equations;  the  former  being  somewhat  more  difficult 
than  those  of  Exercise  24. 

1 .  A  can  do  a  piece  of  work  in  8  days  which  B  can  perform 
in  10  days.  In  how  many  days  can  it  be  done  by  both  working 
together  ? 

Let  X  =  the  number  of  days  required. 

Then,  _  =  the  part  both  can  do  in  one  day. 

X 

Also,  -  =  the  part  A  can  do  in  one  day, 

8 

and  —  =  the  part  B  can  do  in  one  day. 

10 

By  the  conditions,  --\ =  — 

8      10     X 

Clearing  of  fractions,     5  x  +  4  oj  =  40,  or  9  ic  =  40. 

Whence,  x  =  4|,  the  number  of  days  required. 

2.  The  second  digit  of  a  number  exceeds  the  first  by  2 ;  and 
if  the  number,  increased  by  6,  be  divided  by  the  sum  of  its 
digits,  the  quotient  is  5.     Find  the  number. 

Let  X  =  the  first  digit. 

Then,  x-\-2  =  the  second  digit, 

and  2  a;  +  2  =  the  sum  of  the  digits. 

The  number  itself  is  equal  to  10  times  the  first  digit,  plus  the  second. 

Then,     10  a;  +  (x  +  2),  or  11  x  +  2  =  the  number. 

By  the  conditions,        lla;  +  2  +  6  ^  ^ 
2x  +  2 

Whence,  11  x  +  8  =  lOx  +  10,  and  x  =  2. 

Then,  11  x  +  2  =  24,  the  number  required. 


136  ALGEBRA 

3.  Divide  44  into  two  parts  such  that  one  divided  by  the 
other  shall  give  2  as  a  quotient  and  5  as  a  remainder. 

Let  11  =  the  divisor. 

Then,  44  —  ?i  =  the  dividend. 

Now  since  the  dividend  is  equal  to  the  product  of  the  divisor  and 
quotient^  plus  the  remainder,  we  have 

44  —  w  =  2  w  +  5,  whence  —  3  «  =  —  39. 

Then,  w  =  13,  the  divisor, 

and  44  —  w  =  31,  the  dividend. 

4.  Two  persons,  A  and  B,  63  miles  apart,  start  at  the  same 
time  and  travel  towards  each  other.  A  travels  at  the  rate  of 
4  miles  an  hour,  and  B  at  the  rate  of  3  miles  an  hour.  How 
far  will  each  have  travelled  when  they  meet  ? 

Let  4  X  =  the  number  of  miles  that  A  travels. 

Then,  3x  =  the  number  of  miles  that  B  travels. 

By  the  conditions,  4  x  +  3  x  =  63. 

Then,  7  x  =  63,  and  x  =  9. 

Whence,  4  x  =  36,  the  number  of  miles  that  A  travels, 

and  3x  =  27,  the  number  of  miles  that  B  travels. 

It  is  often  advantageous,  as  in  Ex.  4,  to  represent  the  unknown 
number  by  some  multiple  of  x  instead  of  by  x  itself. 

5.  At  what  time  between  3  and  4  o'clock  are  the  hands  of  a 
watch  opposite  to  each  other  ? 

Let  X  =  the  number  of  minute-spaces  passed  over  by  the  minute-hand 
from  3  o'clock  to  the  required  time. 

Then,  since  the  hour-hand  is  15  minute-spaces  in  advance  of  the  minute- 
hand  at  3  o'clock,  X  —  15  —  30,  or  x  —  45,  will  represent  the  number  of 
minute-spaces  passed  over  by  the  hour-hand. 

But  the  minute-hand  moves  12  times  as  fast  as  the  hour-hand. 

Whence,  x  =  12  (x  -  45),  or  x  =  12  x  -  540. 

Then,  -  11  x  =  -  540,  and  x  =  49,Jt, 

Then  the  required  time  is  49^^  minutes  after  3  o'clock. 


FRACTIONAL  AND  LITERAL  LINEAR  EQUATIONS    137 

EXERCISE  62 

1.  The  denominator  of  a  fraction  exceeds  twice  the  numera- 
tor by  4.  If  the  numerator  be  increased  by  14,  and  the  denomi- 
nator decreased  by  9,  the  value  of  the  fraction  is  J.  Find  the 
fraction. 

2.  Divide  197  into  two  parts  such  that  the  smaller  shall  be 
contained  in  the  greater  5  times,  with  a  remainder  23. 

3.  A  piece  of  work  can  be  done  by  A  in  2f  hours,  and  by  B 
in  4i  hours ;  in  how  many  hours  can  the  work  be  done  by  both 
working  together  ? 

4.  The  second  digit  of  a  number  of  two  figures  exceeds  the 
first  by  5 ;  and  if  the  number,  increased  by  6,  be  divided  by 
the  sum  of  the  digits,  the  quotient  is  4.     Find  the  number. 

5.  At  what  time  between  12  and  1  o'clock  are  the  hands  of 
a  watch  opposite  to  each  other  ? 

6.  At  what  time  between  7  and  8  o'clock  is  the  minute-hand 
of  a  watch  10  minutes  in  advance  of  the  hour-hand  ? 

7.  A  piece  of  work  can  be  done  by  A  and  B  working  together 
in  10  days.  After  working  together  7  days,  A  leaves,  and  B 
finishes  the  work  in  9  days.  How  long  will  A  alone  take  to 
do  the  work  ? 

8.  Divide  54  into  two  parts  such  that  twice  the  smaller  shall 
be  3  times  as  much  above  29  as  4  times  the  greater  is  below 
143. 

9.  At  what  time  between  8  and  9  o'clock  are  the  hands  of  a 

watch  together  ? 

10.  The  numerator  of  a  fraction  exceeds  the  denominator  by 
5.  If  the  numerator  be  decreased  by  9,  and  the  denominator 
increased  by  6,  the  sum  of  the  resulting  fraction  and  the  given 
fraction  is  2.     Find  the  fraction. 

11.  At  what  time  between  2  and  3  o'clock  is  the  minute- 
hand  of  a  watch  5  minutes  behind  the  hour-hand  ? 


138  ALGEBRA 

12.  The  second  digit  of  a  number  of  two  figures  is  i  the 
first;  and  if  the  number  be  divided  by  the  difference  of  its 
digits,  the  quotient  is  15,  and  the  remainder  3.  Find  the 
number. 

13.  A  garrison  of  700  men  has  provisions  for  11  days. 
After  3  days,  a  certain  number  of  men  leave,  and  the  pro- 
visions last  10  days  after  this  time.     How  many  men  leave  ? 

14.  A  woman  buys  a  certain  number  of  eggs  for  $  1.05 ;  she 
finds  that  7  eggs  cost  as  much  more  than  18  cents  as  8  eggs 
cost  less  than  27  cents.     How  many  eggs  did  she  buy  ? 

15.  The  width  of  a  field  is  -|  its  length.  If  the  width  were 
increased  by  5  feet,  and  the  length  by  10  feet,  the  area  would 
be  increased  by  400  square  feet.     Find  the  dimensions. 

16.  After  A  has  travelled  7  hours  at  the  rate  of  10  miles  in 
3  hours,  B  sets  out  to  overtake  him,  travelling  at  the  rate  of  9 
miles  in  2  hours.  How  far  will  each  have  travelled  when  B 
overtakes  A  ? 

17.  The  first  digit  of  a  number  of  three  figures  is  f  the 
second,  and  exceeds  the  third  digit  by  2.  If  the  number  be 
divided  by  the  sum  of  its  digits,  the  quotient  is  38.  Find  the 
number. 

18.  A,  B,  and  C  divide  coins  in  the  following  way :  as  often 
as  A  takes  5,  B  takes  4,  and  as  often  as  A  takes  6,  C  takes  7. 
After  the  coins  have  been  divided,  A  has  29  fewer  than  B  and 
C  together.     How  many  coins  were  there  ? 

19.  A  can  do  a  piece  of  work  in  3^  hours,  B  in  3|  hours, 
and  C  in  3f  hours.  In  how  many  hours  can  it  be  done  by  all 
working  together  ? 

20.  A  man  walks  13|-  miles,  and  returns  in  an  hour  less  time 
by  a  carriage,  whose  rate  is  f  as  great  as  his  rate  of  walking. 
Find  his  rate  of  walking. 

21.  At  what  times  between  4  and  5  o'clock  are  the  hands  of 
a  watch  at  right  angles  to  each  other? 


FRACTIONAL  AND  LITERAL   LINEAR  EQUATIONS    139 

22.  A  man  borrows  a  certain  sum,  paying  interest  at  the 
rate  of  5%.  After  repaying  $180,  his  interest  rate  on  the 
balance  is  reduced  to  4^%,  and  his  annual  interest  is  now  less 
by  $  10.80.     Find  the  sum  borrowed. 

23.  The  digits  of  a  certain  number  are  three  consecutive 
numbers,  of  which  the  middle  digit  is  the  greatest,  and  the 
first  digit  the  least.  If  the  number  be  divided  by  the  sum  of 
its  digits,  the  quotient  is  ^-     Find  the  number. 

24.  A  certain  number  of  apples  were  divided  between  three 
boys.  The  first  received  one-half  the  entire  number,  with  one 
apple  additional,  the  second  received  one-third  the  remainder, 
with  one  apple  additional,  and  the  third  received  the  remain- 
der, 7.     How  many  apples  were  there  ? 

25.  A  freight  train  runs  6  miles  an  hour  less  than  a  pas- 
senger train.  It  runs  80  miles  in  the  same  time  that  the 
passenger  train  runs  112  miles.     Find  the  rate  of  each  train. 

26.  A  and  B  each  fire  40  times  at  a  target ;  A's  hits  are  one- 
half  as  numerous  as  B's  misses,  and  A's  misses  exceed  by  15 
the  number  of  B's  hits.  How  many  times  does  each  hit  the 
target  ?  » 

27.  A  freight  train  travels  from  A  to  B  at  the  rate  of  12 
miles  an  hour.  After  it  has  been  gone  3|  hours,  an  express 
train  leaves  A  for  B,  travelling  at  the  rate  of  45  miles  an  hour, 
and  reaches  B  1  hour  and  5  minutes  ahead  of  the  freight. 
Find  the  distance  from  A  to  B,  and  the  time  taken  by  the 
express  train. 

28.  A  tank  has  three  taps.  By  the  first  it  can  be  filled  in 
3  hours  10  minutes,  by  the  second  it  can  be  filled  in  4  hours 
45  minutes,  and  by  the  third  it  can  be  emptied  in  3  hours 
48  minutes.  How  many  hours  will  it  take  to  fill  it  if  all  the 
taps  are  open  ? 

29.  A  man  invested  a  certain  sum  at  3f  %,  and  \^  this  sum 
at  4^%  ;  after  paying  an  income  tax  of  5%,  his  net  annual 
income  is  $  195.70.     How  much  did  he  invest  in  each  way  ? 


140  ALGEBRA 

30.  A  train  leaves  A  for  B,  210  miles  distant,  travelling  at 
the  rate  of  28  miles  an  hour.  After  it  has  been  gone  1  hour 
and.  15  minutes,  another  train  starts  from  B  for  A,  travelling 
at  the  rate  of  22  miles  an  hour.  How  many  miles  from  B  will 
they  meet  ? 

31.  A  can  do  a  piece  of  work  in  |  as  many  days  as  B,  and 
B  can  do  it  in  |-  as  many  days  as  C.  .  Together  they  can  do 
the  work  in  3^^  days.  In  how  many  days  can  each  alone  do 
the  work  ? 

32.  A  vessel  runs  at  the  rate  of  llf  miles  an  hour.  It  takes 
just  as  long  to  run  23  miles  up  stream  as  47  miles  down 
stream.     Find  the  rate  of  the  stream. 

33.  A  man  starts  from  his  home  to  catch  a  train  at  the  rate 
of  one  yard  in  a  second,  and  arrives  2  minutes  late.  If  he  had 
walked  at  the  rate  of  4  yards  in  3  seconds,  he  would  have  been 
3^  minutes  too  early.     Find  the  distance  to  the  station. 

34.  A  crew  has  bread  for  a  voyage  of  50  days,  at  IJ  lb.  each 
a  day.  After  20  days,  7  men  are  lost  in  a  storm,  and  the 
remainder  of  the  crew  have  a  daily  allowance  of  li  lb.  for  the 
balance  of  the  voyage.     Find  the  original  nilmber  of  the  crew. 

35.  A  man  invests  $  230  at  4^  %.  He  then  invests  a  certain 
part  of  a  like  sum  at  3^%,  and  the  balance  at  5J%,  and 
obtains  the  same  income.  How  much  does  he  invest  at  each 
rate  ? 

36.  At  what  times  between  5  and  6  o'clock  do  the  hands  of 
a  watch  make  an  angle  of  45°  ? 

37.  At  a  certain  time  between  12  noon  and  12.30  p.m.,  the 
distance  between  the  hands  is  f  as  great  as  it  is  10  minutes 
later.     Find  the  time. 

38.  A  woman  sells  half  an  egg  more  than  half  her  eggs. 
She  then  sells  half  an  egg  more  than  half  her  remaining  eggs. 
A  third  time  she  does  the  same,  and  now  has  3  eggs  left. 
How  many  had  she  at  first  ? 


FRACTIONAL  AND  LITlJRAL  LINEAR  EQUATIONS    141 

39.  A  merchant  increases  his  capital  annually  by  \  of  itself. 
He  adds  to  his  capital  $  300  at  the  end  of  the  first  year,  and 
$350  at  the  end  of  the  second;  and  finds  at  the  end  of  the 
third  year  that  his  capital  is  |f  of  his  original  capital.  Find 
his  original  capital. 

40.  A  and  B  together  can  do  a  piece  of  work  in  5^  days, 
B  and  C  together  in  6f  days,  and  C  and  A  together  in  5|-  days. 
In  how  many  days  can  it  be  done  by  each  working  alone  ? 

41.  A  fox  is  pursued  by  a  hound,  and  has  a  start  of  77  of 
her  own  leaps.  The  fox  makes  5  leaps  while  the  hound  makes 
4 ;  but  the  hound  in  5  leaps  goes  as  far  as  the  fox  in  9.  How 
many  leaps  does  each  make  before  the  hound  catches  the  fox  ? 

42.  A  man  puts  a  certain  sum  into  a  savings  bank  paying 
4  %  interest.  At  the  end  of  a  year  he  deposits  the  interest, 
receiving  interest  on  the  entire  amount.  At  the  end  of  a 
second  year  and  a  third  year  he  does  the  same,  and  now  has 
$  2812.16  in  the  bank.     What  was  his  original  deposit  ? 

PROBLEMS  IN  PHYSICS 

1.  The  density  of  a  substance  is  defined  as  the  number  of 
grams  in  one  cubic  centimeter.  Hence  the  total  number  of 
grams,  M,  in  any  body  is  equal  to  its  density,  D,  multiplied  by 
its  volume,  F;  or,  to  state  this  relation  algebraically, 

M=DV, 

V  being  given  in  cubic  centimeters,  and  D  in  grams. 

Two  blocks,  one  of  iron  and  one  of  copper,  weigh  the  same 
number  of  grams;  the  iron  has  a  volume  of  10  cubic  centi- 
meters and  a  density  of  7.4 ;  the  copper  has  a  density  of  8.9. 
Find  the  volume  of  the  copper  block. 

2.  When  100  grams  of  alcohol,  of  density  .8,  is  poured  into 
a  cylindrical  vessel,  it  is  found  to  fill  it  to  a  depth  of  10  centi- 
meters. Find  the  area  of  the  base  of  the  cylinder  in  square 
centimeters. 

3.  A  cylindrical  iron  bar,  2  centimeters  in  diameter,  has  a 
mass  of  3  kilograms.     Find  the  length  of  the  bar. 

Let  TT  =  3f 


142  ALGEBRA 

4.  When  a  body  is  weighed  under  water,  it  is  found  to  be 
buoyed  up  by  a  force  equal  to  the  weight  of  the  water  which  it 
displaces. 

If  a  boy  can  exert  a  lifting  force  of  120  pounds,  how  heavy 
a  stone  can  he  lift  to  the  surface  of  a  pond,  if  the  density  of 
stone  is  2.5  and  that  of  water  1  ? 

5.  When  a  straight  bar  is  sup-     g      ^ 2      ^      f 

ported  at  some  point,  o  (Fig.  1), 

and  masses  m^,  m^,  etc.,  are  hung 

from  the  bar  as  indicated  in  the     mi  LJ  fl    m*.  m& 

figure,  it  is  found  that  when  the  "  ^* 

bar  is  in  equilibrium,  the  follow-  ^iQ-  !• 

ing  relation  always  holds, 

mi  X  ao  +  W2  X  bo  =  m^  X  GO -\- m^  X  do -\- m^  X  eo. 

If  a  teeter  board  is  10  feet  long,  where  must  the  support  be 
placed  in  order  that  a  70-pound  boy  at  one  end  may  balance  a 
60-pound  boy  on  the  other  end  plus  a  40-pound  boy  3  feet  from 
the  other  end  ? 

6.  A  bar  40  inches  long  is  in  equilibrium  when  weights  of 
6  pounds  and  9  pounds  hang  from  its  two  ends.  Find  the  posi- 
tion of  the  support. 

7.  If  in  Fig.  1,  ao  =  100,  bo  =  40,  co  =  30,  do  =  60,  eo  =  110, 
and  if  mj  =  40,  m2  =  60,  mg  =  60,  m^  =  15,  and  m^  =  5,  where 
must  a  mass  of  100  be  placed  in  order  to  produce  equilibrium  ? 

8.  A  gas  expands  ^73  ^^  '^^^  volume  at  0°  centigrade  for 
each  degree  of  rise  in  its  temperature ;  i.e.,  the  volume,  F«,  at 
any  temperature,  t,  is  connected  with  the  volume,  Vo,  at  the 
temperature  0°  centigrade  by  the  equation 

or  F.=  Fo(l  +  2kO- 

To  what  volume  will  100  cubic  centimeters  of  air  at  0°  expand 
when  the  temperature  rises  to  50°  centigrade  ? 

9.  To  what  volume  will  100  cubic  centimeters  of  air  at  50° 
centigrade  contract  when  the  temperature  falls  to  0°  centigrade  ? 


FRACTIONAL  AND  LITERAL  LINEAR  EQUATIONS    143 

10.  To  what  volume  will  100  cubic  centimeters  of  air  at  50° 
expand  when  the  temperature  changes  to  75°  ? 

11.  When  a  body  in  motion  collides  with  a  body  at  rest,  the 
momentum  of  the  first  body  (i.e.,  the  product  of  its  mass,  m^,  by 
its  original  velocity,  v^)  is  found  to  be  in  every  case  exactly 
equal  to  the  total  momentum  of  the  two  bodies  after  collision 
(i.e.,  to  the  product  of  the  mass,  mg,  of  the  second  body  times 
the  velocity,  Vg,  which  it  acquires,  plus  the  product  of  rrii  by 
the  velocity,  ^3,  which  it  retains  after  the  collision).  The  alge- 
braic statement  of  this  relation  is 

A  billiard  ball,  the  mass  of  which  is  50  grams,  and  which 
was  moving  at  a  velocity  of  1500  centimeters  a  second,  collided 
with  another  ball  at  rest  which  weighed  30  grams.  In  the 
collision  the  first  ball  imparted  to  the  second  a  velocity  of  1600 
centimeters  per  second.  Find  the  velocity  of  the  first  ball  after 
the  collision. 

PROBLEMS  INVOLVING  LITERAL  EQUATIONS 

162.  Prob.  Divide  a  into  two  parts  such  that  m  times  the 
first  shall  exceed  n  times  the  second  by  b. 

Let  X  =  one  part. 

Then,  a  —  x  =  the  other  part. 

By  the  conditions,  mx  =  n(a  —  x)  +  b. 

mx  =  an  —  nx  +  b, 

mx  ■\-  nx  =  an  +  b. 

x(to  +  7i)  =  an  +  6. 

Whence,  x  =  «!L±_^ ,  the  first  part.  (1) 

m  +  n 

A„j                                „      „      ^      an  +  b     am  +  an  —  an  —  b 
And,  a  —  x=  a —  = ■ — 

m  +  n  m  -\-  n 

am  —  b 


m  -\-  n 


,  the  other  part.  (2) 


144  ALGEBRA 

The  results  can  be  used  as  formuloe  for  solving  any  problem  of  the 
above  form. 

Thus,  let  it  be  required  to  divide  25  into  two  parts  such  that  4  times  the 
first  shall  exceed  3  times  the  second  by  37. 

Here,  a  =  25,  m  =  4,  n  =  3,  and  b  =  37. 

Substituting  these  values  in  (1)  and  (2), 


the  first  part 

^25_ 

X  3  +  37  _ 

7 

_  75  +  37  _ 

7 

112  _ 

7 

=  16, 

and  the  second  part 

-25. 

X4-37 

7 

_  100  -  37  . 

7 

_63_ 

7  " 

=  9. 

EXERCISE  63 

1.  Divide  a  into  two  parts  whose  quotient  shall  be  m. 

2.  If  A  can  do  a  piece  of  work  in  m  hours,  and  A  and  B 
together  in  n  hours,  in  how  many  hours  can  B  alone  do  the 
work  ? 

3.  Divide  a  into  two  parts  such  that  the  sum  of  one-mth  the 

first  and  one-r<th  the  second  shall  equal  b. 

10 

4.  A  courier  who  travels  a  miles  a  day  is  followed  by  another 

who  travels  Similes  a  day.     How  maij^  days  must  the  second 
start  after  the  first  to  overtake  him  after  c  days  ? 

5.  Divide  a  into  three  parts  such  that  the  first  shall  be 
one-mth  the  second  and  one-nth  the  third. 

6.  The  length  of  a  field  is  m  times  its  width.  If  the  length 
were  increased  by  a  feet,  and  the  width  by  b  feet,  the  area 
would  be  increased  by  c  square  feet.  Find  the  dimensions  of 
the  field. 

7.  A  courier  who  travels  a  miles  a  day  is  followed  after  b 
days  by  another.  How  many  miles  a  day  must  the  second 
courier  travel  to  overtake  the  first  after  c  days  ? 

8.  If  A  can  do  a  piece  of  work  in  a  hours,  B  in  5  hours,  C 
in  c  hours,  and  D  in  d  hours,  how  many  hours  will  it  take  to 
do  the  work  if  all  work  together  ? 


FRACTIONAL   AND  LITERAL   LINEAR  EQUATIONS    145 

9.  A  vessel  can  be  filled  by  two  taps  in  a  and  h  minutes, 
respectively,  and  emptied  by  a  third  in  c  minutes.  How  many 
minutes  will  it  take  to  fill  the  tank  if  all  the  taps  are  open  ? 

10.  Divide  a  into  two  parts  such  that  one  shall  be  m  times 
as  much  above  h  as  the  other  lacks  of  c. 

11.  A  can  do  a  piece  of  work  in  one-mth  as  many  days  as  B, 
and  B  can  do  it  in  one-Tith  as  many  as  C.  If  they  can  do  the 
work  in  p  days,  working  together,  in  how  many  days  can  each 
alone  do  the  work  ? 

12.  A  was  m  times  as  old  as  B  a  years  ago,  and  will  be  n 
times  as  old  as  B  in  6  years.     Find  their  ages  at  present. 

13.  How  many  minutes  after  n  hours  after  12  o'clock  will 
the  hands  of  a  watch  be  together  ? 

14.  A  and  B  together  can  do  a  piece  of  work  in  a  hours, 
B  and  C  together  in  h  hours,  and  A,  B,  and  C  together  in  c 
hours.     In  how  many  hours  can  each  alone  do  the  work  ? 

15.  How  many  minutes  after  2  o'clock  will  the  minute-hand 
of  a  watch  be  n  minutes  in  advance  of  the  hour-hand  ? 

16.  A  and  B  together  can  do  a  piece  of  work  in  m  days, 
B  and  C  together  in  n  days,  and  C  and  A  together  in  p  days. 
How  many  days  will  it  take  to  do  the  work  if  all  work 
together  ? 

17.  A  sum  of  money,  amounting  to  m  dollars,  consists 
entirely  of  quarters  and  dimes,  there  being  n  more  dimes  than 
quarters.     How  many  are  there  of  each  ? 


146  ALGEBRA 

XIL     SIMULTANEOUS   LINEAR  EQUATIONS 

CONTAINING  TWO  OR  MORE  UNKNOWN  NUMBERS 

163.  An  equation  containing  two  or  more  unknown  numbers 
is  satisfied  by  an  indefinitely  great  number  of  sets  of  values  of 
these  numbers. 

Consider,  for  example,  the  equation  x-\-y  =  5. 

Putting  x  =  l,  we  have  1  -f-  ?/  =  5,  or  ?/  =  4. 

Putting  x  =  2,  we  have  2  -{-y=  5,  or  y  =  S;  etc. 

Thus  the  equation  is  satisfied  by  the  sets  of  values 

x  =  l,y  =  4:, 

and  x  =  2,  y  =  3;  etc. 

An  equation  which  is  satisfied  by  an  indefinitely  great  num- 
ber of  sets  of  values  of  the  unknown  numbers  involved,  is 
called  an  Indeterminate  Equation. 

•    164.   Consider  the  equations 

f       ^  +  2/  =  5,  (1) 

[2x  +  2y=10.  (2) 

Equation  (1)  can  be  made  to  take  the  form  of  (2)  by  multi- 
plying both  members  by  2 ;  then,  every  set  of  values  of  x  and 
y  which  satisfies  one  of  the  equations  also  satisfies  the  other. 

Such  equations  are  called  equivalent. 

Again,  consider  the  equations 

|^4-2/  =  5,  (3) 

[x-^y  =  3.  (4) 

In  this  case,  it  is  not  true  that  every  set  of  values  of  x  and  y 
which  satisfies  one  of  the  equations  also  satisfies  the  other; 
thus,  equation  (3)  is  satisfied  by  the  set  of  values  x  =3,  y  =  2, 
which  does  not  satisfy  (4). 

If  two  equations,  containing  two  or  more  unknown  numbers, 
are  not  equivalent,  they  are  called  Independent. 


SLMULTANEOUS  LINEAR  EQUATIONS  147 

165.  Consider  the  equations 

x  +  y  =  5y  (1) 

x-\-y  =  6.  (2) 

It  is  evidently  impossible  to  find  a  set  of  values  of  x  and  y 
which  shall  satisfy  both  (1)  and  (2). 
Such  equations  are  called  Inconsistent. 

166.  A  system  of  equations  is  called  Simultaneous  when  each 
contains  two  or  more  unknown  numbers,  and  every  equation 
of  the  system  is  satisfied  by  the  same  set,  or  sets,  of  values 
of  the  unknown  numbers ;  thus,  each  equation  of  the  system 

(x  +  y  =  5, 
\x~y  =  S, 
is  satisfied  by  the  set  of  values  x  —  4:,y  =  l. 

A  Solution  of  a  system  of  simultaneous  equations  is  a  set  of 
values  of  the  unknown  numbers  which  satisfies  every  equation 
of  the  system ;  to  solve  a  system  of  simultaneous  equations  is 
to  find  its  solutions. 

167.  Two  independent  simultaneous  equations  of  the  form 
ax-{-by  =  c  may  be  solved  by  combining  them  in  such  a  way  as 
to  form  a  single  equation  containing  but  one  unknown  number. 

This  operation  is  called  Elimination. 

ELIMINATION  BY  ADDITION  OR  SUBTRACTION 

168.  1.    Solve  the  equations 

Multiplying  (1)  by  4, 

Multiplying  (2)  by  3, 

Adding  (3)  and  (4), 

Whence, 

Substituting  x  =  2  in  (1), 

Whence, 

The  above  is  an  example  of  elimination  by  addition. 


'5x-3y=::19. 

(1) 

,7x-\-4.y=    2. 

(2) 

20  a:  -  12  ?/  =  76. 

(3) 

21x  +  12y=    6. 

(4) 

41  X  =  82. 

(5) 

x=    2. 

(6) 

10  -  3  ?/  =  19. 

(7) 

—  Sy  =  9,  or'y  =  - 

-3. 

(8) 

15x  +  Sy=       1. 

(1) 

10x-7y  =  -24:, 

(2) 

SOx  +  iey=       2. 

(3) 

30  a;  -21y=-72. 

(4) 

148  ALGEBRA 

We  speak  of  adding  a  system  of  equations  when  we  mean  placing  the 
sum  of  the  first  members  equal  to  the  sum  of  the  second  members. 

Abbreviations  of  this  kind  are  frequent  in  Algebra  ;  thus  we  speak  of 
multiplying  an  equation  when  we  mean  multiplying  each  of  its  terms. 

2.    Solve  the  equations 

Multiplying  (1)  by  2, 

Multiplying  (2)  by  3, 

Subtracting  (4)  from  (3),  S7  y  =  7-4,  and  y  =  2. 

Substituting  y  =  2  in  (1),  15  a;  +  16  =  1. 

Whence,  15  x  =  —  15,  and  x  =  —  1. 

The  above  is  au  example  of  elimination  by  subtraction. 

From  the  above  examples,  we  have  the  following  rule : 

If  necessary,  multiply  the  given  equations  by  such  numbers  as 
will  make  the  coefficients  of  one  of  the  unknown  numbers  in  the 
resulting  equations  of  equal  absolute  value. 

Add  or  subtract  the  resulting  equations  according  as  the  coeffi- 
cients of  equal  absolute  value  are  of  unlike  or  like  sign. 

If  the  coefficients  which  are  to  be  made  of  equal  absolute  value  are 
prime  to  each  other,  each  may  be  used  as  the  multiplier  for  the  other 
equation  ;  but  if  they  are  not  prime  to  each  other,  such  multipliers  should 
be  used  as  will  produce  their  lowest  common  multiple. 

Thus,  in  Ex.  1,  to  make  the  coefficients  of  y  of  equal  absolute  value, 
we  multiply  (1)  by  4  and  (2)  by  3 ;  but  in  Ex.  2,  to  make  the  coefficients 
of  X  of  equal  absolute  value,  since  the  L.C.M.  of  10  and  15  is  30,  we  mul- 
tiply (1)  by  2  and  (2)  by  3. 

EXERCISE  64 

In  several  examples  in  the  following  set  other  letters  than  a?*] 
and  y  are  used  to  represent  unknown  numbers. 
Solve  by  the  method  of  addition  or  subtraction : 

•   l4i»-f    2/  =  14.  '  l7a;-f-42/  =  23. 

■   l3a;-82/  =  -35.  "  1    51/+    9a;  =  -23. 


SIMULTANEOUS  LINEAR  EQUATIONS 


149 


ilx- 
[Sx- 
|15a; 
I    6x 


32/=    10. 
5y  =  -5. 


11. 


'■1 


15a;-f    Sy  =  3. 
12y  =  5. 
10x-\-15y  =  -22. 
7x-\-20y  =  -    4. 


{>•{ 


6x-\-lly  =  31. 
6y  —  11  X  =  74. 
9u  +  6v  =  -16. 
lSu-{-7v  =  -22. 


19. 


13    |12a^-ll2/  = 

{12y-llx  =  -21 


8.1   *— 82/  =  -   3.               i4_. 

1 11  a;  +  5  2/ =  —  15. 

24p-    7^  = 

52. 

.18i>  +  13i  =  - 

-34. 

1    9aj-142/  =  -30.                      f  19a; +  202/  = 
*   l21a;  +  132/=      67.                   '  l21a;  +  162/  = 

-35. 

-57. 

•^    fl3?n-7w  =  15.                   _    ri2a;  +  ll2/  = 
10.   1                                                16.  1 

I    8m-47i=    9.                          l28aj-172/  = 

172. 
60. 

ELIMINATION  BY  SUBSTITUTION 

169.   Ex.     Solve  the  equations  \  ^ 

l82/-5ic  =  -17. 

(1) 

(2) 

Transposing  —  5  a;  in  (2),                             8  y  =  5  x  —  17. 

Whence,                                                           y  =  ^^-^^. 

8 

(3) 

Substituting  in  (1),           .    7a;  -  9^^^  ~  ^^)  =  15. 

(4) 

Clearing  of  fractions,           56ic-9(5ic-17)  =  120. 

Or,                                          66  a;  -  45  X  +  153  =  120. 

Uniting  terms,                                             11  a;  =  —  33. 

Whence,                                                           a;  =  —  3. 

(5) 

Substituting  x  =  -  3  in  (3),     y=  ~  ^^  ~ 

iI  =  -4. 

(6) 

From  the  above  example,  we  have  the  following  rule : 

From  one  of  the  given  equations  find  the  value  of  one  of  the 
unknown  numbers  in  terms  of  the  other,  and  substitute  this  value 
in  place  of  that  number  in  the  other  equation. 


'  V 

EXERCISE  65 

Solve  by  the  method  of  substitution 

r    x  +  2y=:ll: 
•   [3x-}-5y  =  29. 

9. 

^8e-3/=47. 

.6e-7/=21. 

(2x-\-    y=   8. 
'   [7x-4.y  =  4.3. 

10. 

'4.x-lly  =  - 

^9x-^   Sy  = 

71.  ' 
4. 

(5x-6y  =  -9. 
'  l3x-52/  =  -4. 

11.  ^ 

'6x-\-12y  = 
.32/-   4a^  =  - 

41. 
9. 

^    r3.T  +  72/  =  -12. 
*  \9y-Q>x=       1. 

12. 

'7a;+    62/  =  - 
.  9  a;  + 10  2/  =  - 

13. 
11. 

r5|7H-    2r  =  —    4. 

1    8  m  —  15  -?;  = 
'   I12m+    6v  = 

18. 

-11. 

{3x-by=     38. 
'  l3?/-5aj  =  -26. 

|9a;  +  82/  =  57. 
'   l6a;  +  7?/  =  48. 

7. 

25a;-12  2/  =  -19. 

15.  ^ 

'18aj-102/  = 
-15  2/-14a;  = 

29. 

.10a;+    42/  =  -    1. 

-24. 

8. 

.10a;  +  92/  =  -6. 

16. 

7^-92/  =  - 
.  llfl;  +  4  2/  =  - 

22. 

■89. 

ELIMINATION 

BY  COM 

PARISON 

■  2  a; 
170.   Ex.     Solve  the  equations   ■ 

-5y  =  -16. 
-\-Ty=      5. 

(1) 
(2) 

Transposing  —  by  in  (1), 

2x=5 

y-lQ- 

"Whence, 

x=^ 

y-16 

2 

(3) 

Transposing  7  ?/  in  (2), 

3x  =  5 

-72/. 

Whence, 

x.=  ^ 

3 

(4) 

Equa 

ting  values  of  x^ 

2 

-7y, 
3 

(5) 

SIMULTANEOUS  LINEAR  EQUATIONS 


151 


Clearing  of  fractions, 

15  y  _  48  =  10  -  14  y. 

Transposing, 

29  y  =  58. 

Whence, 

y  =  2. 

(6) 

Substituting  y  =  2  in  (3), 

.  =  l«-l«.-3. 

(7) 

From  the  above  example,  we  have  the  following  rule : 

From  each  of  the  given  equations,  find  the  value  of  the  same 
unknoivn  number  in  terms  of  the  other,  a7id  place  these  values 
equal  to  each  other. 

EXERCISE  66 

Solve  by  the  method  of  comparison : 


I 


3. 


6. 


8. 


2x-\-3y  =  U. 
ic  -f  4  2/  =  17. 
6x-    2/=     27. 
Sy-3x  =  -36. 
3x-\-2y  =  -31. 
7x-3y  =  -34:. 
5iB-8?/  =  -46. 
2x-\-3y  =  -   6. 
5  x  —  2y=   4. 
4:x  +  7y  =  29. 
jr-3s  =  -lS. 
4r  — 5s  =  —   7. 
3x-Sy  =  -13. 
6»  —  4?/  =  —   5. 
807-   7y  =  -   6. 
6x  +  lly  =  -37. 


9. 


10. 


11. 


12. 


13. 


14. 


9x-{-5y  =  —   1. 
9y-6x=     13. 

S  x-\-    5y  =  o. 
12iK  +  102/  =  7. 
10/1+   6k  =  -15. 
Uk-15h  =  -58. 

Sx+  9?/=  8. 
10a?  +  12  2/  =  ll. 
lld  +  16^  =  64. 

7  d  - 12  ^  =  13. 

Sx-7y  =  -68. 
16y-5x  =  S9. 
(9x-   6y  =  -17. 
'  l7a74-15?/  =  -46. 


16. 


-\-ioy 
8  a;  +  5  2/  =  49. 
13x-h6y  =  S6. 


171.  If  the  given  equations  are  not  in  the  form  ax-^by  =  c, 
they  should  first  be  reduced  to  this  form,  when  they  may  be 
solved  by  either  method  of  elimination. 


152  ALGEBRA 

\— —         =0.  (1) 

1.  Solve  the  equations  j  a; -I- 3     y-\-4:  ^^ 

[x{y-2)-y(x-5)  =  -13,  (2) 

Multiplying  each  term  of  (1)  by  (x  +  3)(2/  +  4), 

7  2/  +  28  -  3  X  -  9  =  0,  or  7  y  -  3  X  =  -  19.  (3) 

From  (2),     xy -2x-xij  +  6y  =  -lS,  or  by -2x  =  -lS.  (4) 

Multiplying  (3)  by  2,  Uy-Qx  =  -  38.  (5) 

Multiplying  (4)  by  3,  15  2,  _  6  x  =  -  39.  (6) 

Subtracting  (5)  from  (6),  y  =:  _    1. 

Substituting  in  (4) ,  _  5  _  2  x  =  -  13. 

Whence,  -2x  =  —   8,  or  x  =  4. 

In  solving  fractional  simultaneous  equations,  we  reject  any 
solution  which  does  not  satisfy  the  given  equations. 

{2x-\-Sy         =13.  (1) 

2.  Solve  the  equations  ]      1  1     _   ^  ,^, 

[x-2     y^~      '  ^"^ 

Multiplying  each  term  of  (2)  by  (x  —  2)  (y  —  3),  we  have 

?/  -  3  4-  X  -  2  =  0,  or  y  =  -  X  +  6.  (3) 

Substituting  in  (1),  2  x  -  3  x  +  15  =  13,  or  x  =  2. 

Substituting  in  (3),  ?/  =  -  2  +  5  =  3. 

This  solution  satisfies  the  first  given  equation,  but  not  the  second  ;  then 
it  must  be  rejected. 

y  EXERCISE  67 

Solve  the  following : 


1.  \ 


2«_5j^_l 
5        6  2' 

X ■ 5y_5 
6     T-2* 


3. 


rila;-3y^3a;4-y 
2.  11  8      *  4. 

[   Sx-5y  =  l, 


'  4:X  —  Sy     X 
14 

9 

-1. 

2x 

+  32/  =  - 

-10 

e-2t-8 

2 

7* 

5e  +  2t  = 

-7. 

SIMULTANEOUS  LINEAR  EQUATIONS 
11.  ' 


153 


5. 


\     2  5 


8. 


9. 


10. 


11. 


x-{-5     y-\-l 


=  0. 


6.  \ 


X ^ =  5. 


11 

9     x  +  5 
2         3 


-3  2/. 


y    r  (2  a;  - 1)(2/ -  4)  -  (oj  -  5)(2  2/ +  5)  =  121. 
l  4a;-32/  =  -29. 


8 


07  —  3 


9 


y-5 
5 


^2aj-l      32/4-4 
'  cc  4- 11      ?/  —  6 


=  0. 
=  0. 

-4. 


10 


7 

x-1 

2 

c?-2n     ^_1. 
3cZ  +  /i  +  3         5* 

c;  +  3  7i    ^7 
(^+4  7i_7      11 ' 

.08a;  +  .9i/  =  .048. 
.3  a; -.35  2/ =.478. 


12. 


13. 


-{ 


15. 


'  2x—3y     'ix-^-Qy  ^  _1 
4       "*"       3       ~     2 
5x-{-2y     7y-3x^S9 
2       "^       5  10* 


a^-f2/^       1  . 
cc  — 2/  10 

3a;  +  8^6.T-l 
2/-4  ~22/  +  3* 

5a;-i(3a5-22/  +  5)=ll. 
|(a^-4  2/)-K«^-2/)  =  16. 


'8 

x- 
4 

^3 

,,^, 

6 

aj 

2/-7 
5 

5 
8 

16. 


17. 


18. 


\6x±By_3x-^^  ^ 

16  5  ^ 


5x  —  4:y_ 

5x-\-4:y 


13 


2fl;  +  5y4-l      3x4-2/-3_     ^,  o,,     o 


a;  —  4  2/  4-  6 
8  a; -2  2/ -18 


154 


19. 


20. 


3x-y-l- 


ALGEBRA 
1 


3x  +  5  2y- 


=  0. 


.     i(3^-l-..)-i(-|)=^- 


5x-7y  +  2     3x  —  4:y-j-7 


2/4-4. 


21. 


22. 


23. 


7x  —  3y-\-4.      ex  —  5y-\-7  _        r> 
4  5 

5^_2^     cc     3  ?/ 
12 ^_3 2__ 

7  23     ~ 

4  2 

U-2  2/4-4(2a;-2/)=0. 

'21»  +  1      5v  +  2        63.^-130v 


3 


21 


3a;  +  l      2x-{-y     x  +  2y    ^^ 
7  2  8 

4  a;  — 2      5x-{-Ay  _x  —  y 


2a;  +  3      3a;  +  4 


2  5 

17 


=  0. 


24.  \4:y-\-5,    62/  +  7     2(4 2/ +  5) (6 2/ +  7) 

t  (a^_l)(2/  +  2)-(a;  +  3)(2/4-4)  =  12. 


25. 


[  :08^±35  __  :13x±^  ^  .32  ^  +  .17  2,  +  .21. 
.15  .6 

.02  a; +  .17      .08  y  -  .47 


.9 


0. 


172.   Solution  of  Literal  Simultaneous  Equations. 

In  solving  literal  simultaneous  linear  equations,  the  method 
of  elimination  by  addition  or  subtraction  is  usually  to  be 
preferred. 


SIMULTANEOUS  LINEAR  EQUATIONS 


155 


Ex.    Solve  the  equations 

Multiplying  (1)  by  b\ 
Multiplying  (2)  by  6, 

Subtracting, 

Whence,  , 

Multiplying  (1)  by  a', 
Multiplying  (2)  by  a, 
Subtracting  (3)  from  (4), 

Whence, 


J   ax  "1-  by  =  c. 
l  a'x  +  h^y  =  c'. 

ah'x  +  Wy  -  h'c. 
a'hx  +  hh'y  =  &c'. 
(a6'  -  a'h)x  =  h'c  -  he'. 

h'c-  he' 
ah'  -  a'h 

aa'x  +  a'hy  =  ca'. 
aa'x  +  ah'y  —  c'a. 


(ah'  —  a'h)y  = 

y  = 


ca  —  ca'. 

c'a  —  ca' 
ah'  -  a'h 


(1) 
(2) 


(3) 
(4) 


In  solving  fractional  literal  simultaneous  equations,  any 
solution  which  does  not  satisfy  the  given  equations  must  be 
rejected.     (Compare  Ex.  2,  §  171.) 


1. 


Solve  the  following 
5x—6y=Sa. 


EXERCISE  68 


4:X+9y=7a 

^  2    raflj+62/  =  l. 
\cx-{-dv  =  l. 

■I 


r  mx—ny 
I  m-x-\-n' 


=  mn. 

-\-n'y=m'n'. 


1/3 


J    8. 


+  dy 
aiX-\-a2y=bi. 
a^  —  ajy=b2. 


5. 


'  2ax  —  by  _ , 
a 


x±by_^^ 
3a+2 


mi     m^     mg 

£.  +  l.  =  i. 
Wi      ^2      % 


7'.  ^' 


m 


n 


7i-\-y     m  —  x 
m  n 


9 


n-\-x 

ai. 


m 


2/ 


V'lO. 


bx—ay=b'^. 

I  {a—b)x  +  by—a^. 

ax  -\-by  =  2  a. 
b^y  =  a'  +  b\ 

(a  +  l)x  -f-  (a  —  2)?/  =  3  a. 
,  (a  +  3)ic  +  (a  —  4)2/  =  7  a. 


l  a^a;  — 


a6(a  —  b)x  +  a6(a  +  &)?/  =  a^  -\-2ab  —  W. 
ax-\-by  =  2. 


156 


ALGEBRA 


•  16. 


12.  \ 


r    m(x  +  2/)  +  n(x  —  y)  =  2. 


13. 


14. 


15. 


I  7nr(x -\-y)  —  n\x  —  y)  —  m—  n. 

(a  +  b)x  -\-  (a  -  h)y  =  2{o}  4-  h^). 
h         _         a 
x  —  a  —  b     y  —  a-\-h 

(a  +  h)x  +  (a  -  6)2/  =  2  c^  -  2  h\ 

y X     _    4  a?) 

a  —  6     a  +  6     a^  —  b^ 

bx-\-ay  =  2. 
.  a6(a  +  6)a;  —  ab{a  —  b)y  =  a^  +  b^,  ' 


y  —  a  +  b  _y  —  a^ 

x  +  a-\-b      x-\-b 
a  —  x      b 


17 


■( 


y  —  b      a 


ay  —  bx=zd^-\.  b^. 
(a  +  b)x-{-{a-b)y  =  2a?-2b\ 


(a  +  b)x  +  (a  —  b)y  =  2  a. 


173.  Certain  equations  in  which  the  unknown  numbers 
occur  in  the  denominators  of  fractions  may  be  readily  solved 
without  previously  clearing  of  fractions. 


,Ex.     Solve  the  equations 


Multiplying  (1)  by  5, 

Multiplying  (2)  by  3, 
Adding, 


10 

_9_ 

:8. 

X 

y 

«+ 

15= 

=-1 

X 

y 

50 

X 

45  _ 

'  y 

40. 

24     45  _ 

x'^  y 

-3. 

—  =  37,  74  =  37  X,  and  x  =  2. 


Substituting  in  (1),  5  -  -  =  8,   -  -  =  3,  and  y  =-S. 
y  9 


(1) 

(2) 


SIMULTANEOUS  LINEAR  EQUATIONS 


157 


EXERCISE  69 


Solve  the  following 


1. 


8. 


X      y 

X     y 

I  X  ^ 


3.  \ 


2.  \ 


+  22/: 


10 
3' 


4. 


ah  _ci  +  h^ 
bx     ay        ah 


h_ 
ax 


a 


x  —  a 
b 


X 


a 
by 

+ 

-  +  - 
a     y 


b^ 


a^b^ 


y  +  'b 


=  1. 


=  1. 


a     b 
a;     y 

\a'      b'       , 
-4--  =  c'. 
[  X      y 

9     11__11 

d'^  s~      2' 


6     21 
id'^  s 


5. 


6. 


-7. 


2 

3a; 

3 

4?/ 

1 

12* 

5 
4a; 

4 

13 

72* 

8_ 

3 
5y 

89 
30 

5 
6x 

_6_ 

2/ 

59 
18 

9.  \ 


p  —  q  _q_  p^  —  2pq  —  q^ 

X         y         p(p-i-q) 


p_p^ 

X 


y 


10. 


-2pq-q^ 
p{p  +  q) 

24 


2x-\-y      x  —  4,y 
7  16 


2x-\-y     x  —  4,y 


-  =  -2. 
=  -3. 


SIMULTANEOUS    LINEAR    EQUATIONS   CONTAINING   MORE 
THAN   TWO    UNKNOWN   NUMBERS 

174.  If  we  have  three  independent  simultaneous  equations, 
containing  three  unknown  numbers,  we  may  combine  any  two 
of  them  by  one  of  the  methods  of  elimination  explained  in 
§§  169  to  171,  so  as  to  obtain  a  single  equation  containing  only 
two  unknown  numbers. 

We  may  then  combine  the  remaining  equation  with  either  of 
the  other  two,  and  obtain  another  equation  containing  the  sa7ne 
two  unknown  numbers. 

By  solving  the  two  equations  containing  two  unknown  num- 
bers, we  may  obtain  their  values ;  and  substituting  them  in 
either  of  the  given  equations,  the  value  of  the  remaining 
unknown  number  may  be  found. 


158  ALGEBRA 

We  proceed  in  a  similar  manner  when  the  number  of  equa- 
tions and  of  unknown  numbers  is  greater  than  three. 

The  method  of  elimination  by  addition  or  subtraction  is 
usually  the  most  convenient. 

In  solving  fractional  simultaneous  equations,  any  solution  which  does 
not  satisfy  the  given  equations  must  be  rejected,     (Compare  Ex.  2,  §  171.) 


Q,X-     4?/  — 

72  =  17. 

(1) 

1.    Solve  the  equations  j    9ic—   ly  — 

16  2  =  29. 

(2) 

lOcc-   hy- 

32  =  23. 

(3) 

Multiplying  (1)  by  3,     \Sx-\2y-2\z  = 

51. 

Multiplying  (2)  by  2,     18 ic  -  14  ^  -  32  z  = 

58. 

Subtracting,                                  2y  +  Uz=- 

-    7. 

(4) 

Multiply ing  ( 1 )  by  5,     SQx-20y-Sbz  = 

85. 

Co) 

Multiplying  (3)  by  3,     SOx-loy  -    9z  = 

69. 

(6) 

Subtracting (5) from  (6)              by-{-26z=- 

-16. 

(7) 

Multiplying  (4)  by  5,                lOy  +  6bz  =■ 

-35. 

Multiplying  (7)  by  2,                I0y  +  52z=- 

-32. 

Subtracting,                                             3z  =■ 

-    3,  or  ^  =  —  1. 

Substituting  in  (7),                        2y  -11=- 

-    7,  or  ?/  =     2. 

Substituting  in  (1),                 6x-S  +  7  = 

17,  or  X  =      3. 

In  certain  cases  the  solution  may  be  abridged  by  aid  of  the 
artifice  which  is  employed  in  the  following  example : 


2.    Solve  the  equations 


lu-\-x  +  y  =  6.  (1) 

x-hy-\-z  =  7.  (2) 

y-{-z  +  u  =  8.  (3) 

_z  +  u-\-x  =  9.  (4) 

Adding,  Su  +  Sx  +  Sy  +  Sz  =  SO. 

Whence,  2t -\- x  + y  +  z  =  10.  (5) 

Subtracting  (2)  from  (5),  '     u  =   S. 

Subtracting  (3)  from  (5),  x=   2. 

Subtracting  (4)  from  (5),  y  =    1. 

Subtracting  (1)  from  (5),  z  =    4. 


SIMULTANEOUS  LINEAR  EQUATIONS 


159 


Solve  the  following : 
(4:x-3y=       1. 

1.  i  4.y-'Sz  =  ~15. 
[4^-30;=  10. 
f  4:X  —  5y  —  6z  = 

2.  \     X—    2/4-    2!  =  - 


EXERCISE  70 


r~.  \ 


4. 


6. 


8. 


22. 
6. 

=      22. 


'< 


l^x  + 

i3x-\-    y—    z=     14.1 
x-\-3y—     z=     16. 
[    x-\-    y—Sz  =  —  10. 
r    g-^    h-k  =  24:. 
l4.g  +  Sh-k  =  61. 
[Qg-5h-k  =  ll. 

i^Sx  +  5y=  1. 
\  9x-\-5z  =  -7. 
[9y-{-3z=  2. 
(5x—    y-{-4:Z  =  —   5. 

I    x-{-Sy-Sz=:-27. 
{2x-5y  =  -26. 
7x  +  6z=-SS. 

3^4 

2/_4      2  +  2' 

2ir  +  4?/—    2!  =  —   2. 
lSx-Sy-\-4:Z  =  -25. 
[10x-\-iy-9z  =  -30. 
3p_l_4g  +  57.  =  10. 
4p-5g-3r  =  25. 
5^_3g_4r  =  21. 


4:X  —  lly  —  5z—    9. 

8a;+4i/—    2  =  11. 
Wx+   7  2/ 4- 6 2;  =  64. 
8X  +  42/+  32  =-52. 
y^l2z  =  -52. 
36. 
j^ll  r    6a;-       2/ +  3  2=     42. 


10. 


11. 


\5x—    2/4-12  2  = 
l9a;4-72/-62  = 


12. 


13. 


14. 


15. 


16. 


nOx-  5y-  z=  2. 
I  6x-172/  +  4  2  =  -46. 
r2a;  +  5  2/+32  =  -7. 

22/- 4 2  =  2-3  X. 
i  5a;+9  2/  =  5  +  7  2. 

r5_8 

a;     2/ 

8_3 

2/      2 

<  '7 


=  -3. 
=  1. 


?5  +  ^  = 
2      3x 

3aj     2/ 

42/     ^ 
152      a; 


ax-{-by  = 
by  -\-  cz  = 
cz  -\-  ax  — 


_3_ 
10' 
7_ 
30' 
J^ 
12' 

g'^  +  b\ 

abc 
W  +  c^ 

abc 
(?-\-a^ 

abc 


160 


ALGEBRA 


17. 


19. 


20. 


21. 


22. 


4:e-12t-20w  =  9. 
8e-    6t-i-10iv  =  5. 


23. 


1 12  e- 18^-    5  IV  =  13. 


24. 


18.     i 


( u  —  x-\-y=     15. 

x  —  y-\-z  =  —  12. 

y  —  z-\-u=:      13. 
I  2;  —  w  -p  a;  =  — 14. 

ri^i    1 

X     y     z 
1^1      1 

y      z      X 


1^11 

iz      X     y 

(ax  —  by  =  a^  —  ab^. 
\  ay  —  hz  =  oj^h  —  If. 


'3u-]-x  = 

-   5. 

4:X  —  y  = 

21. 

5y  +  z  = 

-19. 

6z  —  u  = 

39. 

'-l^l 

=  -3. 

^-M 

31 
12' 

I         3^4 

21 

2* 

ix-y     y 

-z     7 

3 

4         3 

y-z     z 

-{-X 

13 

3      ' 

5 

15 

Z-^X        X- 

2 

-2/_43 
5         10 

25. 


26. 


27. 


28. 


29. 


X      y 

2/      2; 

-  +  -  =  c. 
12      a; 


r&    a 

-  +  -  =  c. 
a;     2/ 

a     c 

2;      a; 

-  4-  -  =  a. 
12/      2; 


0— c     c—a 

J-+-^=6+c4-2a. 
c—a     a— 6 

^  -+T-^=cH-a4-2  6. 


a— 6     6— c 


3  +^=2. 


x+y     x—z 

_6 L_  =  i 

x-hy    y—z 

4  +^=2. 


a;  —  2;     y  —  z 

Sx+  9y-{-15z  =  -29. 
17a;-10  2/+13;2  =  -12. 
lla;-15?/+  7z=     15. 

w  +  3a;  —  22/-2  =  —    3. 

2%-a;-2/  +  3;2=      23. 

u-{-x-\-3y-2z  =  -12. 

[3u-2x-\-y  +  z=      22. 


SIMULTANEOUS  LINEAR  EQUATIONS  161 

(x-{-y  -\-z  =  0. 
30.     \  (b-\-c)x-\-(c-^a)y-{-(a  +  b)z  =  0. 
[  hex  4-  cay  +  ahz  =  1. 
id     v.s_     Q  ( 6x-\-5y     y—4:Z_     14 


31. 


^  +  |-|=   28.  32. 

4      6      3 

6^3     4 


3  5  5 


5x 


7x-\-Sy_     4 
6  9     '  ~     9* 

3       "^     4      ~      2* 


PROBLEMS    INVOLVING    SIMULTANEOUS    LINEAR    EQUA- 
TIONS   WITH    TWO    OR    MORE    UNKNOWN    NUMBERS 

175.  In  solving  problems  where  two  or  more  letters  are 
used  to  represent  unknown  numbers,  we  must  obtain  from 
the  conditions  of  the  problem  as  many  independent  equations 
(§  164)  as  there  are  unknoion  numbers  to  he  deterrairied. 

1.  Divide  81  into  two  parts  such  that  three-fifths  the  greater 
shall  exceed  five-ninths  the  less  by  7. 


Let 

X  =  the  greater  part, 

and 

y  =  the  less. 

By  the  conditions, 

x  +  y  =  Sl, 

(1) 

and 

Sx^6y^ 
6        9  ^ 

(2) 

Solving  (1)  and  (2), 

x  =  45,  y  =  S6. 

2.  If  3  be  added  to  both  numerator  and  denominator  of  a 
fraction,  its  value  is  f ;  and  if  2  be  subtracted  from  both  nu- 
merator and  denominator,  its  value  is  ^ ;  find  the  fraction. 


Let 

n  =  the  numerator, 

and 

d  =  the  denominator. 

By  the  conditions, 

n  +  S     2 
d  +  S     3' 

and 

w-2     1 
d-2     2 

Solving  these  equations, 

n  =  7,  d  =  12;  then,  the  fraction  is  — 

162  •  ALGEBRA 

3.  A  sum  of  money  was  divided  equally  between  a  certain 
number  of  persons.  Had  there  been  3  more,  each  would  have 
received  $  1  less ;  had  there  been  6  fewer,  each  would  have  re- 
ceived $5  more.  How  many  persons  were  there,  and  how 
much  did  each  receive  ? 

Let  X  =  the  number  of  persons, 

and  y  =  the  number  of  dollars  received  by  each. 

Then,  xy  =  the  nuijiber  of  dollars  divided. 

Since  the  sum  of  money  could  be  divided  between  x  +  S  persons,  each 
of  whom  would  receive  y  —  1  dollars,  and  between  x  —  6  persons,  each  of 
whom  would  receive  y  +  6  dollars,  (a;+3)(y  — 1)  and  (a;  — 6)(?/  +  5) 
also  represent  the  number  of  dollars  divided. 

Then,  (x^S)(y-l)=xy, 

and  (x  —  6)  (y  +  5)  =  xy. 

Solving  these  equations,  x  =  12,  y  =  5. 

4.  The  sum  of  the  three  digits  of  a  number  is  13.  If  the 
number,  decreased  by  8,  be  divided  by  the  sum  of  its  second 
and  third  digits,  the  quotient  is  25 ;  and  if  99  be  added  to  the 
number,  the  digits  will  be  inverted.     Find  the  number. 

Let  X  =  the  first  digit, 

y  =  the  second, 
and  z  =  the  third. 

Then,  100  x -}- 10  y -{-  z  =  the  number, 

and  100  z  +  lOy  -\-  x  =  the  number  with  its  digits  inverted. 

By  the  conditions  of  the  problem, 

x  +  y-\-z  =  lS, 

100  X  + 10  y  +  g  -  8  _  ,)^ 

2/  +  -S  ~ 

and  100  X  +  10  y  +  ^;  +  99  =  100 ;?  +  10  ?/  4-  a;. 

Solving  these  equations,  x  =  2,  y  =  S,  z  =  S  ;  and  the  number  is  283. 

5.  A  crew  can  row  10  miles  in  50  minutes  down  stream,  and 
12  miles  in  li  hours  against  the  stream.  Find  the  rate  in 
miles  per  hour  of  the  current,  and  of  the  crew  in  still  water. 


SIMULTANEOUS  LINEAR  EQUATIONS  163 

Let  X  =  number  of  miles  an  hour  of  the  crew  in  still  water, 

and  y  =  number  of  miles  an  hour  of  the  current. 

Then,  x  4-  y  =  number  of  miles  an  hour  of  the  crew  down  stream, 

and  x  —  y  =  number  of  miles  an  hour  of  the  crew  up  stream. 

The  number  of  miles  an  hour  rowed  by  the  crew  is  equal  to  the  dis- 
tance in  miles  divided  by  the  time  in  hours. 

Then,  x  +  ?/  =  10  --  -  =  12,  . 

and  x-y  =  12----  =  S. 

2 

Solving  these  equations,         6  =  10,    y  =  2. 

6.  A  train  running  from  A  to  B  meets  with  an  accident 
which  causes  its  speed  to  be  reduced  to  one-third  of  what  it 
was  before,  and  it  is  in  consequence  5  hours  late.  If  the  acci- 
dent had  happened  60  miles  nearer  B,  the  train  would  have 
been  only  1  hour  late.  Find  the  rate  of  the  train  before  the 
accident,  and  the  distance  to  B  from  the  point  of  detention. 

Let  3  X  =  the  number  of  miles  an  hour  of  the  train  before  the  accident. 

Then,  x  =  the  number  of  miles  an  hour  after  the  accident. 

Let      y  =  the  number  of  miles  to  B  from  the  point  of  detention. 

The  train  would  have  done  the  last  y  miles  of  its  journey  in  -^  hours ; 

3  X 
but  owing  to  the  accident,  it  does  the  distance  in  "  hours. 

X 

Then,  l  =  ^^&,  m 

X     3  X  ^  ^ 

If  the  accident  had  occurred  60  miles  nearer  J5,  the  distance  to  B  from 

the  point  of  detention  would  have  been  y  —  60  miles. 

Had  there  been  no  accident,  the  train  would  have  done  this  in  ^~ 

?/  —  60  ^  * 

hours,  and  the  accident  would  have  made  the  time  ^ hours. 

X 

Then,  y.Il^  =  y^Z^+i,  (2) 

X  3  X 

Subtracting  (2)  from  (1),  ^  =  ^  +  4,  or  —  =  4  ;  whence,  x  =  10. 
X       3x  X 

Then,  the  rate  of  the  train  before  the  accident  was  30  miles  an  hour. 


164  ALGEBRA 


Substituting  in  (1),        io~io'^  ^'  ^^  TK- ^  '  ^^^nce,  y  =  75. 


EXERCISE  71 

1.  Divide  79  into  two  parts  such  that  three-sevenths  the 
less  shall  be  less  by  56  than  four-thirds  the  greater. 

2.  If  the  numerator  of  a  fraction  be  increased  by  4,  the 
value  of  the  fraction  is  f ;  while  if  the  denominator  is  de- 
creased by  3,  the  value  of  the  fraction  is  |.     Find  the  fraction. 

3.  The  sum  of  the  two  digits  of  a  number  is  14 ;  and  if  36 
be  added  to  the  number,  the  digits  will  be  inverted.  Find  the 
number. 

4.  A's  age  is  f  of  B's,  and  15  years  ago  his  age  was  \^  of 
B's.     Find  their  ages. 

5.  If  the  two  digits  of  a  number  be  inverted,  the  quotient  of 
the  number  thus  formed,  increased  by  101,  by  the  original  num- 
ber is  2 ;  and  the  sum  of  the  digits  exceeds  twice  the  excess  of 
the  tens'  digit  over  the  units'  digit  by  5.     Find  the  number. 

6.  If  3  be  added  to  the  numerator  of  a  fraction,  and  7  sub- 
tracted from  the  denominator,  its  value  is  ^ ;  and  if  1  be  sub- 
tracted from  the  numerator,  and  7  added  to  the  denominator, 
its  value  is  f.     Find  the  fraction. 

7.  A's  age  is  twice  the  sum  of  the  ages  of  B  and  C ;  two 
years  ago,  A  was  4  times  as  old  as  B,  and  four  years  ago,  A 
was  6  times  as  old  as  C.     Find  their  ages. 

8.  If  the  greater  of  two  numbers  be  divided  by  the  less,  the 
quotient  is  1,  and  the  remainder  6.  And  if  the  greater,  in- 
creased by  14,  be  divided  by  the  less,  diminished  by  4,  the 
quotient  is  5,  and  the  remainder  4.     Find  the  numbers. 

9.  If  8  yards  of  silk  and  12  yards  of  woolen  cost  $  27,  and 
12  yards  of  silk  and  8  yards  of  woolen  cost  $  28,  find  the  price 
per  yard  of  the  silk  and  of  the  woolen. 

10.  Find  two  numbers  such  that  one  shall  be  n  times  as 
much  greater  than  a  as  the  other  is  less  than  a;  and  the  quo- 
tient of  their  sum  by  their  difference  equal  to  b. 


SIMULTANEOUS  LINEAR  EQUATIONS  165 

11.  A  certain  number  of  two  digits  exceeds  three  times  the 
sum  of  its  digits  by  4.  If  the  digits  be  inverted,  the  sum  of 
the  resulting  number  and  the  given  number  exceeds  three 
times  the  given  number  by  2.     Find  the  number. 

12.  The  sum  of  the  three  digits  of  a  number  is  16 ;  the  digit 
in  the  tens'  place  exceeds  that  in  the  hundreds'  place  by  4 ; 
and  if  297  be  added  to  the  number,  the  digits  will  be  inverted. 
Find  the  number. 

13.  A  rectangular  field  has  the  same  area  as  another  which 
is  6  rods  longer  and  2  rods  narrower,  and  also  the  same  area  as 
a  third  which  is  3  rods  shorter  and  2  rods  wider.  Find  its 
dimensions. 

14.  Find  three  numbers  such  that  the  first  with  one-half  the 
second  and  one-third  the  third  shall  equal  29 ;  the  second  with 
one-third  the  first  and  one-fourth  the  third  shall  equal  28; 
and  the  third  with  one-half  the  first  and  one-third  the  second 
shall  equal  36. 

15.  The  circumference  of  the  large  wheel  of  a  carriage  is  55 
inches  more  than  that  of  the  small  wheel.  The  former  makes 
as  many  revolutions  in  going  250  feet  as  the  latter  does  in 
going  140  feet.  Find  the  number  of  inches  in  the  circumfer- 
ence of  each  wheel. 

16.  If  the  digits  of  a  number  of  three  figures  be  inverted, 
the  sum  of  the  number  thus  formed  and  the  original  number 
is  1615 ;  the  sum  of  the  digits  is  20,  and  if  99  be  added  to  the 
number,  the  digits  will  be  inverted.     Find  the  number. 

17.  A  train  leaves  A  for  B,  112  miles  distant,  at  9  a.m.,  and 
one  hour  later  a  train  leaves  B  for  A;  they  met  at  12  noon. 
If  the  second  train  had  started  at  9  a.m.,  and  the  first  at  9.50 
A.M.,  they  would  also  have  met  at  noon.     Find  their  rates. 

18.  A  boy  has  $  1.50  with  which  he  wishes  to  buy  two  kinds 
of  note-books.  If  he  asks  for  14  of  the  first  kind,  and  11  of  the 
second,  he  will  require  6  cents  more ;  and  if  he  asks  for  11  of 
the  first  kind,  and  14  of  the  second,  he  will  have  6  cents  over. 
How  much  does  each  kind  cost  ? 


166  ALGEBRA 

19.  A  man  invests  $10,000,  part  at  4^%,  and  the  rest  at 
3^%.  He  finds  that  six  years'  interest  on  the  first  invest- 
ment exceeds  five  years'  interest  on  the  second  by  $658. 
How  much  does  he  invest  at  each  rate  ? 

20.  A  man  buys  apples,  some  at  2  for  3  cents,  and  others  at 
3  for  2  cents,  spending  in  all  80  cents.  If  he  had  bought  J 
as  many  of  the  first  kind,  and  f  as  many  of  the  second,  he 
would  have  spent  99  cents.  How  many  of  each  kind  did  he 
buy? 

21.  An  annual  income  of  $800  is  obtained  in  part  from 
money  invested  at  3|^%,  and  in  part  from  money  invested 
at  3%.  If  the  amount  invested  at  the  first  rate  were  invested 
at  3%,  and  the  amount  invested  at  the  second  rate  were  in- 
vested at  3|-%,  the  annual  income  would  be  $825.  How 
much  is  invested  at  each  rate  ? 

22.  A  tank  containing  864  gallons  can  be  filled  by  two 
pipes,  A  and  B.  After  the  pipes  have  been  open  together  for 
9  minutes,  the  pipe  A  is  closed,  and  B  finishes  the  work  of 
filling  in  15 1  minutes.'  If  15  minutes  had  elapsed  before  the 
pipe  A  was  closed,  B  would  have  finished  in  2 J  minutes. 
How  many  gallons  does  each  pipe  fill  in  one  minute  ? 

23.  The  contents  of  one  barrel  is  f  wine,  and  of  another  -| 
wine.  How  many  gallons  must  be  taken  from  each  to  fill  a 
barrel  -whose  capacity  is  24  gallons,  so  that  the  mixture  may 
be  I  wine  ? 

24.  A  boy  spends  his  money  for  oranges.  Had  he  bought 
m  more,  each  would  have  cost  a  cents  less ;  if  n  fewer,  each 
would  have  cost  b  cents  more.  How  many  did  he  buy,  and  at 
what  price  ? 

25.  A  vessel  contains  a  mixture  of  wine  and  water.  If  50 
gallons  of  wine  are  added,  there  is  J  as  much  wine  as  water ; 
if  50  gallons  of  water  are  added,  there  is  4  times  as  much 
water  as  wine.  Find  the  number  of  gallons  of  wine  and  water 
at  first. 


SIMULTANEOUS  LINEAR  EQUATIONS  167 

26.  A  man  buys  15  bottles  of  sherry,  and  20  bottles  of 
claret,  for  ^38.  If  the  sherry  had  cost  j  as  much,  and  the 
claret  ^  as  much,  the  wine  would  have  cost  $38.50.  Find  the 
cost  per  bottle  of  the  sherry,  and  of  the  claret. 

27.  If  a  field  were  made  a  feet  longer,  and  b  feet  wider,  its 
area  would  be  increased  by  m  square  feet ;  but  if  its  length 
were  made  c  feet  less,  and  its  width  d  feet  less,  its  area  would 
be  decreased  by  n  square  feet.     Find  its  dimensions. 

28.  If  the  numerator  of  a  fraction  be  increased  by  a,  and  the 

TYh 

denominator  by  b,  the  value  of  the  fraction  is  — :  and  if  the 

numerator  be  decreased  by  c,  and  the  denominator  by  d,  the  value 

of  the  fraction  is  —     Find  the  numerator  and  denominator, 
m 

29.  A  certain  number  equals  59  times  the  sum  of  its  three 
digits.  The  sum  of  the  digits  exceeds  twice  the  tens'  digit  by 
3  ;  and  the  sum  of  the  hundreds'  and  tens'  digits  exceeds  twice 
the  units'  digit  by  6.     Find  the  number. 

30.  A  piece  of  work  can  be  done  by  A  and  B  in  4|-  hours, 
by  B  and  C  in  2|  hours,  and  by  A  and  C  in  3  hours.  In  how 
many  hours  can  each  alone  do  the  work  ? 

31 .  The  numerator  of  a  fraction  has  the  same  two  digits  as 
the  denominator,  but  in  reversed  order ;  the  denominator  ex- 
ceeds the  numerator  by  9,  and  if  1  be  added  to  the  numerator 
the  value  of  the  fraction  is  |.     Find  the  fraction. 

32.  A  man  walks  from  one  place  to  another  in  5^  hours.  If 
he  had  walked  ^  of  a  mile  an  hour  faster,  the  walk  would  have 
taken  36|  fewer  minutes.  How  many  miles  did  he  walk,  and 
at  what  rate  ? 

33.  A  man  invests  a  certain  sum  of  money  at  a  certain  rate 
of  interest.  If  the  principal  had  been  $  120l)  greater,  and  the 
rate  1  %  greater,  his  income  would  have  been  increased  by 
$  118.  If  the  principal  had  been  $  3200  greater,  aiid  the  rate 
2  %  greater,  his  income  would  have  been  increased  by  $  312. 
What  sum  did  he  invest,  and  at  what  rate  ? 


168  ALGEBRA 

34.  A  sum  of  money  at  simple  interest  amounted  to  $  1868.40 
in  7  years,  and  to  $  2174.40  in  12  years.  Find  the  principal 
and  the  rate. 

35.  A  and  B  together  can  do  a  piece  of  work  in  3|  hours. 
If  A  works  I  as  fast,  and  B  f  as  fast,  they  can  do  it  in  the 
same  time.     In  how  many  hours  can  each  alone  do  the  work  ? 

36.  Two  men  together  can  do  a  piece  of  work  in  30  hours ; 
they  can  also  do  it  if  the  first  man  works  25^  hours,  and.  the 
second  32|  hours.  In  how  many  hours  can  each  alone  do  the 
work  ? 

37.  A  crew  row  IG^  miles  up  stream  and  18  miles  down 
stream  in  9  hours.  They  then  row  21  miles  up  stream  and  19  J 
miles  down  stream  in  11  hours.  Find  the  rate  in  miles  an 
hour  of  the  stream,  and  of  the  crew  in  still  water. 

38.  A  train  travels  from  A  to  B,  228  miles,  and  another 
from  B  to  A.  If  the  trains  start  at  the  same  time,  they  will 
meet  3|  hours  after.  If  the  first  train  starts  3  hours  after  the 
second,  they  will  meet  2  hours  after  the  second  train  starts. 
Find  the  rates  of  the  trains. 

39.  A  man  has  quarter-dollars,  dimes,  and  half-dimes  to  the 
value  of  $  1.40,  and  has  in  all  12  coins.  If  he  replaces  the 
quarters  by  dimes,  and  the  dimes  by  quarters,  the  value  of 
the  coins  would  be  $  1.55.     How  many  has  he  of  each  ? 

40.  The  middle  digit  of  a  number  of  three  figures  is  one-half 
the  sum  of  the  other  two  digits.  If  the  number  be  divided  by 
the  sum  of  its  digits,  the  quotient  is  20,  and  the  remainder  9 ; 
and  if  594  be  added  to  the  number,  the  digits  will  be  invertedo 
Find  the  number. 

41.  A  certain  number  of  workmen  receive  the  same  wages, 
and  receive  together  a  certain  sum.  If  there  had  been  9  more 
men,  and  each  ha'd  received  30  cents  less,  the  total  received 
would  have  been  increased  by  $  12.30.  Had  there  been  8  fewer 
men,  and  each  had  received  40  cents  more,  the  total  received 
would  have  been  decreased  by  f  13.20.  How  many  men  were 
there,  and  how  much  did  each  receive  ? 


SIMULTANEOUS  LINEAR  EQUATIONS  169 

42.  A  merchant  has  three  casks  of  wine,  containing  together 
66  gallons.  He  pours  from  the  first  into  the  second  and  third 
as  much  as  each  of  them  contains  ;  he  then  pours  from  the  sec- 
ond into  the  first  and  third  as  much  as  each  of  them  then  con- 
tains. There  is  now  8  times  as  much  in  the  third  cask  as  in 
the  second,  and  twice  as  much  in  the  first  as  in  the  second. 
How  many  gallons  did  each  have  at  first  ? 

43.  In  a  meeting  of  600  persons,  a  measure  is  defeated  by  a 
certain  majority.  It  is  afterwards  successful  by  double  this 
majority,  and  the  number  of  persons  voting  for  it  is  |  as  great 
as  the  number  voting  against  on  the  former  occasion.  How 
many  voted  for,  and  how  many  against,  the  measure  on  the 
former  occasion? 

44.  I  bought  apples  at  3  for  5  cents,  and  oranges  at  2  for  5 
cents,  spending  in  all  $  1.70.  I  sold  three-fourths  of  the  apples 
and  one-half  of  the  oranges  for  $  1.10,  and  made  a  profit  of  5 
cents  on  the  latter  transaction.    How  many  did  I  buy  of  each  ? 

45.  A  gives  to  B  and  C  as  much  money  as  each  of  them  has; 
B  then  gives  to  A  and  C  as  much  money  as  each  of  them  then 
has ;  C  then  gives  to  A  and  B  as  much  money  as  each  of  them 
then  has.     Each  has  now  $  8.     How  much  had  each  at  first  ? 

46.  A  has  one-half  as  many  dimes  as  dollars,  and  B  eight- 
sevenths  as  many  dimes  as  dollars.  They  have  together  3 
more  dollars  than  dimes,  and  B's  money  is  60  cents  less  than 
A's.     How  much  money  has  each  ? 

47.  A  man  buys  a  certain  number  of  $  100  railway  shares, 
when  at  a  certain  rate  per  cent  discount,  for  ^  1050 ;  and  when 
at  a  rate  per  cent  premium  twice  as  great,  sells  one-half  of  them 
for  $  1200.     How  many  shares  did  he  buy,  and  at  what  cost  ? 

48.  A  and  B  can  do  a  piece  of  work  in  |4  hours,  A  and  C 
in  J/  hours,  A  and  D  in  |-i  hours,  and  B  and  C  in  ^-  hours. 
How  many  hours  will  it  take  each  alone  to  do  the  work  ? 

49.  A  and  B  run  a  race  of  280  feet.  The  first  heat,  A  gives 
B  a  start  of  70  feet,  and  neither  wins  the  race.  The  second 
heat,  A  gives  B  a  start  of  35  feet,  and  beats  him  by  6|  seconds. 
How  many  feet  can  each  run  in  a  second  ? 


170  ALGEBRA 

50.  A,  B,  C,  and  D  play  at  cards.  After  B  has  won  one- 
half  of  A's  money,  C  one-third  of  B's,  D  one-fourth  of  C's,  and 
A  one-fifth  of  D's,  they  have  each  $  10,  except  B,  who  has  $  16. 
How  much  had  each  at  first  ? 

51.  The  sum  of  the  four  digits  of  a  number  is  14.  The 
sum  of  the  last  three  digits  exceeds  twice  the  first  by  2. 
Twice  the  sum  of  the  second  and  third  digits  exceeds  3  times 
the  sum  of  the  first  and  fourth  by  3.  If  2727  be  added  to 
the  number,  the  digits  will  be  inverted.     Find  the  number. 

52.  A  and  B  run  a  race  of  210  yards.  The  first  heat,  A 
gives  B  a  start  of  8  seconds,  and  beats  him  by  20  yards.  The 
second  heat,  A  gives  B  a  start  of  70  yards,  and  is  beaten  by 
2  seconds.     How  many  yards  can  each  run  in  a  second  ? 

53.  A  sum  of  money  consists  of  half-dollars,  dimes,  and 
half-dimes.  Its  value  is  as  many  dimes  as  there  are  pieces  of 
money;  and  its  value  is  also  as  many  half-dollars  as  there  are 
dimes  less  1.  The  number  of  dimes  is  5  more  than  the  num- 
ber of  half-dollars.     Find  the  number  of  each  coin. 

54.  The  fore-wheel  of  a  carriage  makes  a  revolutions  more 
than  the  hind-wheel  in  travelling  b  feet.  If  the  circumference 
of  the  fore- wheel  were  m  times  as  great,  and  the  circumference 
of  the  hind-wheel  n  times  as  great,  the  fore- wheel  would  make 
c  revolutions  more  than  the  hind-wheel  in  travelling  d  feet. 
Find  the  circumference  of  each  wheel. 

55.  A  train  running  from  A  to  B  meets  with  an  accident 
which  delays  it  a  hours.  It  then  proceeds  at  a  rate  one-nth  of 
its  former  rate,  and  arrives  at  J5  6  hours  late.  Had  the  acci- 
dent occurred  c  miles  nearer  B,  the  train  would  have  been  d 
hours  late.  Find  the  rate  of  the  train  before  the  accident, 
and  the  distance  to  B  from  the  point  of  detention. 

56.  A  man  buys  60  shares  of  stock,  each  having  the  par 
value  $100,  part  paying  dividends  at  the  rate  of  3|%,  and  the 
remainder  at  the  rate  of  4^%.  If  the  first  part  had  paid  divi- 
dends at  the  rate  of  4^%,  and  the  other  at  the  rate  of  3|%, 
the  total  annual  income  would  have  been  $12  less.  How  many 
shares  of  each  kind  did  he  buy  ? 


SIMULTANEOUS  LINEAR  EQUATIONS  171 

176.  Interpretation  of  Solutions. 

1.  The  length  of  a  field  is  10  rods,  and  its  breadth  8  rods; 
how  many  rods  must  be  added  to  the  breadth  so  that  the  area 
may  be  60  square  rods  ? 

Let  X  =  number  of  rods  to  be  added. 

By  the  conditions,  10  (8  +  ic)  =  60. 

Then,  80  +  10a;  =  60,  or  x= -2. 

This  signifies  that  2  rods  must  be  subtracted  from  the  breadth  in  order 
that  the  area  may  be  60  square  rods.     (Compare  §  16.) 

If  we  should  modify  the  problem  so  as  to  read  : 

"  The  length  of  a  field  is  10  rods,  and  its  breadth  8  rods ;  how  many 
rods  must  be  subtracted  from  the  breadth  so  that  the  area  may  be  60 
square  rods  ?" 
and  let  x  denote  the  number  of  rods  to  be  subtracted,  we  should  find  x  =  2. 

A  negative  result  sometimes  indicates  that  the  problem  is 
impossible. 

2.  If  11  times  the  number  of  persons  in  a  certain  house, 
increased  by  18,  be  divided  by  4,  the  result  equals  twice  the 
number  increased  by  3 ;  find  the  number. 

Let  X  =  the  number. 

By  the  conditions,       '^'^^-^^^  =  2 x  +  3. 
4 

Whence,  llcc  + 18  =  8x  + 12,  and  a;  =  -2. 

The  negative  result  shows  that  the  problem  is  impossible. 

A  problem  may  also  be  impossible  when  the  solution  is 
fractional. 

3.  A  man  has  two  kinds  of  money :  dimes  and  cents.  The 
total  number  of  coins  is  23,  and  their  value  37  cents.  How 
many  has  he  of  each  ? 

Let  X  =  number  of  dimes. 

Then,  23  —  x  =  number  of  cents. 

The  X  dimes  are  worth  10  x  cents. 


172  ALGEBRA 


14 

Then,  by  the  conditions,  10  x  +  23  —  aj  =  37  ;  and  x  =  — 

The  fractional  result  shows  that  the  problem  is  impossible. 


EXERCISE  72 

Interpret  the  solutions  of  the  following : 

1.  If  the  length  of  a  field  is  12  rods,  and  its  width  9  rods, 
how  many  rods  must  be  subtracted  from  the  width  so  that  the 
area  may  be  144  square  rods  ? 

2.  A  is  44  years  of  age,  and  B  12  years;  how  many  years 
ago  was  A  3  times  as  old  as  B  ? 

3.  The  number  of  apple  and  pear  trees  in  an  orchard  is  23 ; 
and  7  times  the  number  of  apple  trees  plus  twice  the  number 
of  pear  trees  ecjuals  82.     How  many  are  there  of  each  kind  ? 

4.  The  number  of  silver  coins  in  a  purse  exceeds  the  num- 
ber of  gold  coins  by  3,  and  5  times  the  number  of  silver  coins 
exceeds  3  times  the  number  of  gold  coins  by  3.  How  many 
are  there  of  each  kind  ? 

5.  A's  assets  are  double  those  of  B.  When  A  has  gained 
$  250,  and  B  $  170,  A's  assets  are  5  times  those  of  B.  Find 
the  assets  of  each. 

6.  A  cistern  has  two  pipes.  When  both  are  open,  it  is  filled 
in  7^  hours ;  and  the  first  pipe  alone  can  fill  it  in  3  hours. 
How  many  hours  does  the  second  pipe  take  to  fill  it  ? 

7.  The  numerator  of  a  fraction  is  4  times  the  denominator ; 
and  if  the  numerator  be  diminished  by  9,  and  the  denominator 
by  15,  the  value  of  the  fraction  is  |.     Find  the  fraction. 

8.  A  and  B  are  travelling  due  east  at  the  rates  of  4i-  and  3| 
miles  an  houi^  respectively.  At  noon  A  is  5  miles  due  east  of 
B.  How  many  miles  to  the  east  of  A's  position  at  noon  will 
he  overtake  B  ? 

'9.  A  has  $  720,  and  B  $  300.  After  A  has  gained  a  certain 
sum,  and  B  has  gained  two-thirds  this  sum,  A  has  3  times  as 
much  money  as  B.     How  much  did  each  gain  ? 


GRAPHICAL   REPRESENTATION 


173 


XIII.     GRAPHICAL  REPRESENTATION 
177.   Rectangular  Co-ordinates  of  a  Point. 


Lm 


^ 


M 


m. 


A. 


i/i 


X 

Let  XX'  and  YY'  be  straight  lines  intersecting  at  right 
angles  at  0;  let  Pi  be  any  point  in  the  plane  of  XX'  and  YY', 
and  draw  line  Pi^i  perpendicular  to  XX'. 

Then,  OMi  and  M^Pi  are  called  the  rectangular  co-ordinates, 
or  simply  the  co-ordinates,  of  Pi;  OM^  is  called  the  abscissa, 
and  MiPi  the  ordinate. 

178.  It  is  understood,  in  the  definitions  of  §  177,  that 
abscissas  measured  to  the  right  of  0  are  positive,  and  to  the 
left,  negative;  also,  that  ordinates  measured  upwards  from  XX' 
are  positive,  and  downwards,  negative.  ^ 

Thus,  let  P2  be  to  the  left  of  YY,  and  above  XX',  and  P3  and 
P4  below  XX',  respectively  to  the  left  and  right  of  YY,  an^" 
draw  lines  P^M^,  P^M^,  and  P^M^  perpendicular  to  XX'. 

Let         0Mi  =  5,     MM  =  3,     M,0  =  5,     0M,  =  2, 
JfiPi  =  3,   M,P2  =  5,    PsMs=S,    P,M,  =  4.. 

Then,  the  abscissa  of  Pi  is  -f  5,  and  its  ordinate  +  3 
the  abscissa  of  Po  is  —  3,  and  its  ordinate  -f  5 
the  abscissa  of  P.  is  —  5,  and  its  ordinate  —  3 
the  abscissa  of  P.  is  4-  2,  and  its  ordinate  —  4. 


174 


ALGEBRA 


179.  The  lines  of  reference,  XX'  and  YT',  are  called  the 
axis  of  X,  and  axis  of  T,  respectively ;  and  0  the  origin. 

We  express  the  fact  that  the  abscissa  of  a  point  is  b,  and  its 
ordinate  a,  by  saying  that,  for  the  point  in  question,  x  =  b  and 
y  =  a;  or,  more  concisely,  we  speak  of  the  point  as  the  point 
(b,  a)  ;  where  the  first  term  in  parentheses  is  understood  to  be 
the  abscissa,  and  the  second  term  the  ordinate. 

If  a  point  lies  upon  XX',  its  ordinate  is  zero ;  and  if  it  lies 
upon  YY',  its  abscissa  is  zero. 

The  co-ordinates  of  the  origin  are  (0,  0). 

180.  Plotting  Points. 

To  plot  a  point  when  its  co-ordinates  are  given,  lay  off  the 
abscissa  to  the  right  or  left  of  0,  according 
as  it  is  H-  or  — ,  and  then  draw  a  perpen- 
dicular, equal  in  length  to  the  ordinate, 
above  or  below  XX',  according  as  the 
ordinate  is   -|-  or  — . 

Thus,  to  plot  the  point  (—2,  3),  lay  off 
2  units  to  the  left  of  O  upon  XX',  and 
then  erect  a  perpendicular  3  units  in  length  above  XX'. 


^ 


-2,3)- 


Y' 


■^ 


EXERCISE  73 


Plot  the  following  points : 


1.  (1,  4). 

2.  (2,  -2). 

3.  (-3,  6). 

4.  (-2,  -4). 

5.  (3,  1). 


6.  (-4,  -3). 

7.  (-1,  2). 

8.  (4,  -6). 

9.  (r,  3). 
10.  (-6,  1). 


11.  (5,  0). 

12.  (0,  4). 

13.  (-2,  0). 

14.  (0,  -3). 


GRAPH    OF   A   LINEAR    EQUATION    INVOLVING   TWO 
UNKNOWN   NUMBERS 


181.  Consider  the  equation  y  =  x  +  2. 


GRAPHICAL   REPKESENTATION 


175 


If  we  give  any  numerical  value  to  x,  we  may,  by  aid  of  the 
relation  y  =  0:-{-  2,  calculate  a  corresponding  value  for  y. 


If  a;  =  0, 
If  a;  =  1, 

If  a;  =  2, 


2/ -2. 


(-1) 
2/  =  3.    .  (B) 

2/ =  4.       .  (C) 

If  a;  =  3,        2/  =  5.  (D) 

If  a^=-l,     2/  =  l.  (E) 

If  a;  =  -  2,     2/  =  0.  (F) 

Ifaj  =  _3,     y  =  -l;  etc.      ((9) 

Now  let  these  be  regarded  as  the  co-ordinates  of  points ;  and 
let  the  points  be  plotted,  as  explained  in  §  180. 

Thus,  to  plot  the  point  A^  lay  oif  2  units  above  0  upon  YY'. 

The  points  will  be  found  to  lie  on  a  certain  line,  GD,  which 
is  called  the  Graph  of  the  given  equation. 

By  assuming  fractional  values  for  x,  we  may  obtain  intermediate 
points  of  the  graph. 

EXERCISE  74 

Eind  by  the  above  method  the  graphs  of  the  following 
equations : 

1.  2/  =  2aj  +  3.  3.   4?/  +  a^  =  6.  5.   y=:5x, 

2.  2/  =  -3a;-4.       4.   3y-2x  =  -12.      6.   3x-^2y  =  0. 

182.  AVe  shall  always  find  (and  it  can  be  proved)  that  a 
linear  equation,  involving  two  unknown  numbers,  has  a  straight 
line  for  a  graph. 

Then,  since  a  straight  line  is  determined  by  any  two  of  its 
points,  it  is  sufficient,  when  finding  the  graph  of  a  linear  equa- 
tion involving  two  unknown  numbers,  to  find  two  of  its  points, 
and  draw  a  straight  line  through  them. 

The  points  most  easily  determined  are  those  in  which  the 
graph  intersects  the  axes. 

For  all  points  on  OX,  y  =  0;  hence,  to  find  where  the  graph 
cuts  OX,  put  ?/  =  0,  and  calculate  the  value  of  x. 

To  find  where  the  graph  cuts  OY,  put  a;  =  0,  and  calculate 
the  value  of  y. 


176 


ALGEBRA 


Ex.     Plot  the  graph  of  2cc  +  3?/  =  —  7. 

7 


Put  ?/  =  0 ;  then  2  x  =  —  7,  and  x 


2 


Then  plot  A  on   OX',  -  units  to  the  left 
ofO.         •  2  X- 

Put  X  =  0  ;  then  3 y  =  —  7  and  y  =-*-. 

Then  plot  5  on  OT',  -  units  below  0. 

3 
Draw  the  straight  line  AB  \  this  is  the  re- 
quired graph. 

The  above  method  cannot,  of  course,  be  used  for  a  straight  line  passing 
through  the  origin,  nor  for  the  equations  of  §  183. 

183.   Consider  the  equation  y  =  5. 

This  means  that  every  point  in  the  graph 
has  its  ordinate  equal  to  5. 

Then  the  graph  is  the  straight  line  AB, 
parallel  to  XX',  and  5  units  above  it. 

In  like  manner,  the  graph  of  x  =  —  3  is 
the  straight  line  CD,  parallel  to  YY',  and 
3  units  to  the  left  of  it. 

The  graph  of  y  —  0  is  the  axis  of  X,  and  the  graph  of  a;  =  0  is  the 
axis  of  Y. 

EXERCISE  75 

Plot  the  graphs  of  the  following  equations : 

1.  3x-^2y  =  6.         3.   x  =  2,  5.  16x-27y:=-72. 

2.  a;-42/  =  4.  4.   y  =  -4:.  6.     Sx-{-Wy  =  -6. 


c 

Y 

^^Tf 

■^ 

A 

0 

" 

D 

Y' 

INTERSECTIONS   OF  GRAPHS 

184.   Consider  the  equations 

p-    y  =  -5.     (AB) 
U  +  3y=     3.     (CD) 

Let  AB  be  the  graph  of  x—y=—5,    x- 
and  CD  the  graph  oi  x-{-3y  =  3. 


GRAPHICAL  REPRESENTATION 


177 


Let  AB  and  CD  intersect  at  E. 

Since  E  lies  on  each  graph,  its  co-ordinates  must  satisfy  both 
given  equations ;  hence,  to  find  the  co-ordinates  of  E,  we  solve 
the  given  equations. 

In  this  case  the  solution  is  a;  =  —  3,  y  =  2;  and  it  may  be 
verified  in  the  figure  that  these  are  the  co-ordinates  of  E. 

We  then  have  the  following  important  principle : 

If  the  graphs  of  tivo  linear  equations,  with  two  unknown  num-' 
hers,  intersect,  the  co-ordinates  of  the  point  of  intersection  form  a 
solution  of  the  equations  represented  by  the  graphs. 

EXERCISE  76 

Verify  the  principle  of  §  184  in  the  following  equations : 


■1 


4  aj  +  5  2/  =      24. 


Sx 


2    (3a^  +  72/  = 
[Sx  +  3y  = 


2y  =  -    5. 

5. 

18. 


3.  i 


f  5  ic  —  4  2/  = 


0. 


[Tx-{-6y  =  -29.  ■ 

9x-\-14:y  =  -25. 

y=     22. 


r  9  05  + 14  2/  = 


As  additional  examples,  the  pupil  might  verify  graphically 
the  solutions  of  Exs.  3,  8,  11,  and  12,  Exercise  65,  and  of 
Exs.  7,  8,  9,  and  16,  Exercise  66. 


185.   Graphs  of  Inconsistent  Linear  Equations  with  Two  Un- 
known Numbers. 

Consider  the  equations 

(Sx-2y=     5.     (AB) 
[6x-4.y  =  -7.     {CD) 

The  first  equation  can  be  put  in  the  form 
6x  —  4:y=10,  by  multiplying  both  mem- 
bers by  2. 

Then,  the  given  equations  are  inconsistent 
(§  165),  and  it  is  impossible  to  find  any  values 
of  X  and  y  which  satisfy  both  equations. 


178 


ALGEBRA 


We  shall  always  find  that  two  inconsistent  equations,  with 
two  unknown  numbers,  are  represented  by  parallel  graphs;  for 
if  the  graphs  could  intersect  at  any  point,  the  co-ordinates  of 
this  point  would  be  a  solution  of  the  given  equations  (§  184). 

186.  Graphs  of  Indeterminate  Linear  Equations  with  Two  Un- 
known Numbers. 

Consider  the  equations 

(3x-2y=   5. 
[6x-4:y  =  10. 


The  first  equation  can  be  put  in  the  form 
of  the  second,  by  multiplying  both  members 
by  2,  and  the  graphs  coincide. 

The  given  equations  are  not  independent 
(§  164)  ;  in  any  similar  case,  we  shall  find  that  the  graphs  are 
coincident. 

EXERCISE  77 

Verify  the  principles  of  §§  185  and  186  in  the  following 
equations: 

(3x-\-4.y=     12.  ^    f2x-   7y  =  U. 


2. 


3x-{-4.y  =  -12. 
2x-   5y=   0. 
6x-15y  =  30. 


'■I 


4a;-14?/  =  28. 
5x-\-   6  2/ =  15. 
15x-\-lSy  =  4.5. 


187.   Graphical  Representation  of  Linear  Expressions  involving 
One  Unknown  Number. 

Consider  the  expression  3x  +  5. 

Put  2/  =  3  a?  +  5  ;  and  let  the  graph  of 
this  equation  be  found  as  in  §  183,  /B 

Putting  y  =  0,  x  =  —  -^;  then  the  graph 
cuts  XX'  I  units  to  the  left  of  0. 

Putting  x  =  0,  y  =  5;  then  the  graph 
cuts  YY'  5  units  above  0. 

The  graph  is  the  straight  line  AB. 


GRAPHICAL  REPRESENTATION"  179 

188.   Graphical  Representation  of  Roots  of  Equations  (§  81). 

In  order  to  find  the  abscissa  of  the  point  A  (§  187),  where 
the  graph  of  3  a?  +  5  intersects  XX',  we  solve  the  equation 
3x  +  o  =  0(§  182). 

That  is,  the  abscissa  of  ^  is  a  root  of  the  equation  3  a;+5=0. 

Hence,  the  abscissa  of  the  point  in  which  the  graph  of  the  first 
member  of  any  linear  equation,  with  one  unknown  number ^  inter- 
sects XX',  is  the  root  of  the  equation. 

EXERCISE  78 

Plot  the  graphs  of  the  first  members  of  the  following  equa- 
tions, and  in  each  case  verify  the  principle  of  §  188 : 

1.   2a;  +  7  =  0.  2.   5aj-4  =  0. 


( 


180  ALGEBRA 

XIV.    INEQUALITIES 

189.  The  Signs  of  Inequality,  >  and  < ,  are  read  "  is  greater 
than  "  and  "  is  less  than,"  respectively. 

Thus,  a  >  &  is  read  "  a  is  greater  than  6  "  ;  a  <  6  is  read  "  a 
is  less  than  6." 

190.  One  number  is  said  to  be  greater  than  another  when 
the  remainder  obtained  by  subtracting  the  second  from  the 
first  is  a  positive  number. 

One  number  is  said  to  be  less  than  another  when  the  remain- 
der obtained  by  subtracting  the  second  from  the  first  is  a 
negative  number. 

Thus,  if  a  —  6  is  a  positive  number,  a  >  6 ;  and  if  a  —  &  is  a 
negative  number,  a<b. 

191.  An  Inequality  is  a  statement  that  one  of  two  expres- 
sions is  greater  or  less  than  another. 

The  First  Member  of  an  inequality  is  the  expression  to  the 
left  of  the  sign  of  inequality,  and  the  Second  Member  is  the 
expression  to  the  right  of  that  sign. 

Any  term  of  either  member  of  an  inequality  is  calle4  a  term 
of  the  inequality. 

Two  or  more  inequalities  are  said  to  subsist  in  the  same  sense 
when  the  first  member  is  the  greater  or  the  less  in  both. 

Thus,  a >  6  and  G>d  subsist  in  the  same  sense. 

PROPERTIES   OF  INEQUALITIES 

192.  An  inequality  will  continue  in  the  same  sense  after  the 
same  number  has  been  added  to,  or  subtracted  from,  both 
members. 

For  consider  the  inequality  a'>b. 
By  §  190,  a  —  6  is  a  positive  number. 


INEQUALITIES  181 

Hence,  eaxih  of  the  numbers 

(a  4-  c)  —  (5  +  c),  and  (a  —  c)  —  (6  —  c) 
is  positive,  since  each  is  equal  to  a  —  b. 

Therefore,      a-\-c>b-\-c,  and  a  —  c>b  —  c.  (§  190) 

193.  It  follows  from  §  192  that  a  term  may  be  transposed 
from  one  member  of  an  inequality  to  the  other  by  changing  its 
sign. 

If  the  same  term  appears  in  both  members  of  an  inequality,  affected 
with  the  same  sign,  it  may  be  cancelled. 

194.  If  the  signs  of  all  the  terms  of  an  inequality  be  changed 
the  sign  of  inequality  must  be  reversed.  ^ 

For  consider  the  inequality   a  —  b>c  —  d. 

Transposing  every  term,        d  —  c>b  —  a.  (§  193) 

That  is,  b  —  a<d  —  c, 

195.  An  inequality  will  continue  in  the  same  sense  after  both 
members  have  been  multiplied  or  divided  by  the  same  positive 
number. 

For  consider  the  inequality  a  >  b. 

By  §  190,  a  —  5  is  a  positive  number. 

Hence,  if  m  is  a  positive  number,  each  of  the  numbers 

m(a  —  b)  and     ~   ,    or   ma  —  mb  and ,  is  positive. 

m  mm 

Therefore,  ma  >  mb,  and  —  >  —  • 

m      m 

196.  It  follows  from  §§  194  and  195  that  if  both  members  of 
an  inequality  be  multiplied  or  divided  by  the  same  negative  num- 
ber, the  sign  of  inequality  must  be  reversed. 

197.  If  any  number  of  inequalities,  subsisting  in  the  same 
sense,  be  added  member  to  member,  the  residting  inequality  will 
also  subsist  in  the  same  sense. 


182  ALGEBRA 

For  consider  the  inequalities  a>h,  a^  > b',  a" >  b",  •••. 
Each  of  the  numbers,  a  —  b,  a'  —  b',  a"  —  b",  •••,  is  positive. 

Then,  their  sum  a-b +  a'  -b'  -j-a"  -b"  + -•-, 

or,  .  a-\-a' -{-a" -{ (6  +  6' +  6"+ •••)> 

is  a  positive  number. 

Whence,       a-^a'  +  a"  +  "•>b +  b' +  b"  +  •••• 

If  two  inequalities,  subsisting  in  the  same  sense,  be  subtracted  member 
from  member,  the  resulting  inequality  does  not  necessarily  subsist  in  the 
same  sense. 

Thus,  if  a  >  &  and  a'  >  6',  the  numbers  a  —  b  and  a'  —  b'  are  positive. 

But  (a  —  b)-(a'  -  b'),  or  its  equal,  (a  -  a')-(b  -  b'),  may  be  posi- 
tive, negative,  or  zero  ;  and  hence  a  —  a'  may  be  greater  than,  less  than, 
or  equal  to  6  —  b'. 

198.  If  a  >  6  and  a'  >b',  and  each  of  the  numbers  a,  a',  b, 
6',  is  positive,  then  ^^,  ^  ^^,^ 

Since  a'  >  b',  and  a  is  positive, 

aa'>ab'  (§195).  (1) 

Again,  since  a>b,  and  b'  is  positive, 

ab'>bb\  (2) 

From  (1)  and  (2),  aa'  >  66'. 

199.  If  we  have  any  number  of  inequalities  subsisting  in 
the  same  sense,  as  a>6,  a'  >6',  a">  b",  •••,  and  each  of  the 
numbers  a,  a',  a",  •-,  6,  6',  6",  •••,  is  positive,  then 

aa'a"...>66'6".... 

For  by  §198,  aa'>66'. 

Also,  a">6". 

Then  by  §198,  aaV  >66'6". 

Continuing  the  process  with  the  remaining  inequalities,  we 
obtain  finally 


INEQUALITIES  183 

200.  Examples. 

1.  Find  the  limit  of  x  in  the  inequality 

Multiplying  both  members  by  3  (§  195),  we  have 

21  X  -  23  <  2  «+  15. 
Transposing  (§  193),  and  uniting  terms, 

19  X  <  38. 
Dividing  both  members  by  19  (§  195), 

x<2. 
[This  means  that,  for  any  value  ofic<2,7x-^<^  +  5.  J 

2.  Find  the  limits  of  x  and  y  in  the  following : 

r3a;  +  22/>37.  (1) 

l2a;  +  32/  =  33.  (2) 

Multiply  (1)  by  3,  9  x  +  6  y  >  111. 

Multiply  (2)  by  2,  4x  +  6y=    m. 

Subtracting  (§192),  5x>   45,  and  x>9. 

Multiply  (1)  by  2,  6  x  -f  4  ?/  >     74. 

Multiply  (2)  by  3,  6  x  +  9  y  =      99. 

Subtracting,  —  5  ?/  >  —  25. 

Divide  both  members  by       —  5,  ?/  <  5  (§  196). 

(This  means  that  any  values  of  x  and  y  which  satisfy  (2),  also  satisfy 
(1),  provided  x  is  >  9,  and  y  <  5.) 

3.   Between  what  limiting  values  of  a;  is  a^  —  4  a;  <  21  ? 
Transposing  21,  we  have 

x2  -  4  X  is  <  21,  if  x2  -  4  X  -  21  is  <  0. 

That  is,  if  (x  +  3)(x  -  7)  is  negative. 

Now  (x  +  3)  (x  -  7)  is  negative  if  x  is  between  -  3  and  7  ;  for  if  x  is 
<  -  3,  both  X  +  3  and  x  -  7  are  negative,  and  their  product  positive  ;  and 
if  X  is  >  7,  both  x  +  3  and  x  —  7  are  positive. 

Hence,  x^  —  4  x  is  <  21,  if  x  is  >-  3,  and  <  7. 


184  ALGEBRA 

EXERCISE  79 

Find  the  limits  of  x  in  the  following : 

1.  (4a;+5)2-4<(8a;  +  5)(2a;  +  3). 

2.  (3  a;  +  2)  (ic  +  3)  -  4  aj  >  (3  a;  -  2)(a;  -  3)  +  36. 

3.  (x-\-4:)(5x-2)-\-(2x-  3)2  >  (3  a;  +  4)^  -  78. 

4.  (x  -  S)(x  +  4)(a;  -  5)  <  (a;  +  l)(x  -  2){x  -  3). 

5.  a'(x-l)<2  b''(2 x-1)  - ab,  if  a-2b  is  positive. 

6.  ^~^  +  2  >^"^^,  if  m  and  %  are  positive,  and  m<^n. 

n  m 

Find  the  limits  of  x  and  y  in  the  following : 


7  a;  —  4  2/  >  41. 
l3a;4-T2/  =  35. 


y      r5a;  +  62/<45.  g^ 

\Sx-4.y  =  -ll. 

9.   Find  the  limits  of  x  when 

3a;-ll<24-lla;,  and  5  aj  +  23<20 x  +  3. 

10.  If  6  times  a  certain  positive  integer,  plus  14,  is  greater 
than  13  times  the  integer,  minus  63,  and  17  times  the  integer, 
minus  23,  is  greater  than  8  times  the  integer,  plus  31,  what  is 
the  integer  ? 

11.  If  7  times  the  number  of  houses  in  a  certain  village, 
plus  33,  is  less  than  12  times  the  number,  minus  82,  and  9 
times  the  number,  minus  43,  is  less  than  5  times  the  number, 
plus  61,  how  many  houses  are  there  ? 

12.  A  farmer  has  a  number  of  cows  such  that  10  times  their 
number,  plus  3,  is  less  than  4  times  the  number,  plus  79 ;  and 
14  times  their  number,  minus  97,  is  greater  than  6  times  the 
number,  minus  5.     How  many  cows  has  he? 

13.  Between  what  limiting  values  of  a;  is  ar^H-3a;<4? 

14.  Between  what  limiting  values  of  a;  is  x^<Sx  — 15. 

15.  Between  what  limiting  values  of  a;  is  3  a;^  + 19  a;<  —20  ? 

201.   If  a  and  b  are  unequal  numbers, 
a^-{-b^>2ab. 


INEQUALITIES  185 

For  (a-by>0',  or,  cc" -2  ab +  b^>0. 

Transposing  —2ab,  a^ -\- b^  >  2  ab. 

1.  Prove  that,  if  a  does  not  equal  3, 

(a4-2)(a-2)>6a-13. 
By  the  above  principle,  if  a  does  not  equal  3, 

a2  -f  9  >  6  a. 
Subtracting  13  from  both  members, 

(j2_4;>6a_13,  or  (a  +  2)(a  -  2)  >6  «- 13. 

2.  Prove  that,  if  a  and  b  are  unequal  positive  numbers. 

a^-\-b^>a'b  +  b'a. 
We  have,  a^  +  b'^>2  ah,  or  ^2  _  ab  -\-b^>  ab. 

Multiplying  both  members  by  the  positive  number  a  +  6, 

a^+b^>  a%  +  b'^a. 

EXERCISE  80 

1.  Prove  that  for  any  value  of  x,  except  {, 

3a;(3a;-10)>-25. 

2.  Prove  that  for  any  value  of  x,  except  |, 

4x(x-5)>8a;-49. 

3.  Prove  that  for  any  values  of  a  and  6,  if  4  a  does  not  equal 
^^'  (4  a  +  3  6) (4  a  -  3  6)  >  6  &  (4  a  -  3  6). 

4.  Prove  that  for  any  values  of  x  and  y,ii  5x  does  not  equal 
^^'  5 x(5 x- 6 y)  >2 y(5 x-S y). 


Prove  that,  if  a  and  6  are  unequal  positive  numbers, 

5.   a%  +  ab'>2a'b'.  6.    ?H-->2. 

b     a 

7.   a^-^a'b-\-ab^-{-b^>2ab{a-\-b). 


186  ALGEBRA 


^^  XV.    INVOLUTION 

202.  Involution  is  the  process  of  raising  an  expression  to  any 
power  whose  exponent  is  a  positive  integer. 

We  gave  in  §  96  a  rule  for  raising  a  monomial  to  any  power 
whose  exponent  is  a  positive  integer. 

203.  Any  Power  of  a  Fraction. 


Wehave,      gy=?x?x^     axaxa     a- 


b      b      b      bxbxb      W' 


and  a  similar  result  holds  for  any  positive  integral  power  of  -  • 

Then,  a  fraction  may  be  raised  to  any  power  whose  exponent  is 
a  positive  integer  by  raising  both  numerator  and  denominator  to 
the  required  power. 

^      /     2  x'Y         (2  x'\\.  ...  (2  x'Y  32  a^    ,.  qan 


EXERCISE  81 

Find  the  values  of  the  following : 
^    r6a%^y  •   o    /     3aV^^ 


/     2  m'xy 
V       nhf  J  ' 


7  MJ  \       Wy 

o    /9  mny  .    (_±^y  6    f  ^'^'  \ 

\Sp'  J'  ■  V     5y'zy'  '  [Sb'cny 

204.   Square  of  a  Polynomial. 
We  find  by  actual  multiplication : 

a  -\-b  +  c 
a  -{-b  -\-c 
a^  +   ab  -i-     ac 

+   ab  +6^+    be 

4-     ac         -{-     bc-\-c^ 

a^-\-2ab-{-2ac-\-b^  +  2bc  +  (f 


INVOLUTION  187 

The  result,  for  convenience  of  enunciation,  may  be  written : 

(a  +  6  +  c)2  =  a-  +  &'  +  c'  +  2a6  +  2ac  +  2  6c. 
In  like  manner  we  find : 

(a-f-6  +  c  +  (^)2  =  a2  +  &2  +  c2  +  d2 

-{-2ab-{-2ac  +  2ad-{-2bc-^2bd  +  2cd', 
and  so  on. 

We  then  have  the  following  rule : 

The  square  of  a  polyyiomial  is  equal  to  the  sum  of  the  squares 
of  its  terms,  together  with  twice  the  product  of  each  term  by  each 
of  the  following  terms. 

Ex.    Expand  (2x''-^x- B)\ 

The  squares  of  the  terms  are  4  x*,  9  sc^,  and  25. 

Twice  the  product  of  the  first  term  by  each  of  the  following  terms  gives 
the  results  -  12  x^  and  -  20  x^. 

Twice  the  product  of  the  second  term  by  the  following  term  gives  the 
result  30  x. 

Then,        (2  «2  -  3  «  -  5)2  =  4  a^4  +  g  3^2  +  25  -  12  jc^  -  20  ic^  _^  30  a; 

=  4x*-12a:3-llx2  +  30a;  +  25. 

EXERCISE  82 
Square  each  of  the  following : 

\.   a—b  +  c.  10.  x^  —  4:xy  —  5y\ 

2.  x-\-y-z.  11.  6a^-\-ab-Sb^ 

3.  n^-3n-l.  12.  2a^P-8a  +  9. 

4.  3x-{-y  +  2z.  13.  6  a^^ - 4 a.^  +  5 2/*. 

5.  l-\-Sx-4:a^.  14.  a-^b-c-d. 

6.  2a'"- 5a"- 1.  15.  a- 6  + c  +  d 

7.  44-3m3  +  2m«.  16.  a^  +  a^  +  a-3. 

8.  77i3_n2  +  6.  '  17.  2a;3-4a^-3£c-f-l. 

9.  2x  +  3y-\-5z.  18.  3-2a  +  4a2-5a^ 


188  ALGEBRA 


19. 

m  +  4- 

.1                        20.   ^f     «  +  - 

m                                    3       X     c 

205. 

Cube  of  a  Binomial. 

We  find  by 

actual  multiplication : 

a  +b 

a^  +  2a'b-{-    ab^ 

a'b-\-2ab'-\-b^ 

aa? 


That  is,  the  cube  of  the  sum  of  two  numbers  is  equal  to  the 
cube  of  the  first,  plus  three  times  the  square  of  the  first  times  the 
second,  plus  three  times  the  first  times  the  square  of  the  second, 
plus  the  cube  of  the  second. 

Again,  (a  -by  =  a^-2ab  +  b^ 

a  -b 


a^-2a^b-\-     ab^ 
-    a^b-^2ab^-b^ 


(a-by  =  a'-3a'b-{-3ab'-b^ 

That  is,  the  cube  of  the  difference  of  two  numbers  is  equal  to 
the  cube  of  the  first,  minus  three  times  the  square  of  the  first 
times  the  second,  plus  three  times  the  first  times  the  square  of  the 
second,  minus  the  cube  of  the  second. 

1.  Find  the  cube  oia  +  2b. 

We  have,    (a  +  2  6)3  =  a^  +  3  a%2  6)  -f  3  a(2  6)2  +  (2  6)8 
=  a8  +  6  a26  +  12  a62  +  8  6». 

2.  Find  the  cube  oi2o^-5y\ 

(2  x8  -  5  2/2)3  =  (2  a:3)3  _  3(2  a:3)2(5  y^)  +  3(2  aj8)(5  y2)2_  (5  ^2)8 
=  8  x9  -  60  xhj^  +  150  x^y^  -  125  y^ 

The  cube  of  a  trinomial  may  be  found  by  the  above  method, 
if  two  of  its  terms  be  enclosed  in  parentheses;  and  regarded 
as  a  single  term. 


INVOLUTION  189 

3.   Find  the  cube  of  a^  -  2  a;  - 1. 

(x2-2a;-l)8  =  [(x2-2a;)-  ip 

=  (x2  -  2  x)3  -  3(^2  -  2  x)2  +  3(a:2  _  2  a;)  -  1 
=,afi  -  6x^  +  12  X^-Sx^-  3(x*  -  4  ic3  +  4 x2)  +  3(x2  _  2 x)  -1 
=  a;6  -  6  x5  +  12  cc*  -  8  x3  -  3  x*  +  12  x3  -  12  a:2  4-  3  x2  -  6  X  -  1 
=  x6-6x5  +  9x4  +  4x3-9x2-6x-l. 

EXERCISE  83 
Cube  each  of  the  following : 

1.  a^b-ab\                  7.   2a-{-6x.  14.  a-b-\-c. 

2.  a 4- 3.                        8.   5m^-3n^.  15.  1  — a  — al 

3.  2x  +  y.                    9.   3a-6a2.  16.  a  +  264-c. 

4.  a-5  6.                    10.   2a^6  +  7c^  17.  x'-hx-S. 

5.  6a^+l.                   11.   8a2"»  +  3a.  18.  3-n  +  2n2. 

6.  m-47i3.                 12.   9x'-4.y\  19.  2a2  +  3a-4. 

1 Q        'ni'^         2  71 
*'^'      O 2  ' 

2n     m^ 


190  ALGEBRA 


t      XVI.    EVOLUTION 

206.  If  an  expression  when  raised  to  the  nth.  power,  n  being 
a  positive  integer,  is  equal  to  another  expression,  the  first 
expression  is  said  to  be  the  nth  Root  of  the  second. 

Thus,  if  a"  =  &,  <x  is  the  nth  root  of  b. 

Evolution  is  the  process  of  finding  any  required  root  of  an 
expression. 

207.  The  Radical  Sign,  ^,  when  written  before  an  expres- 
sion, indicates  some  root  of  the  expression. 

Thus,   Va  indicates  the  second,  or  square  root  of  a; 
Va  indicates  the  third,  or  cube  root  of  a ; 
Va  indicates  the  fourth  root  of  a ;  and  so  on. 

The  index  of  a  root  is  the  number  written  over  the  radical 
sign  to  indicate  what  root  of  the  expression  is  taken. 

If  no  index  is  expressed,  the  index  2  is  understood. 

An  even  root  is  one  whose  index  is  an  even  number ;  an  odd 
root  is  one  whose  index  is  an  odd  number. 

EVOLUTION  OF  MONOMIALS 

208.  We  will  now  show  how  to  find  any  root  of  a  monomial, 
which  is  a  perfect  power  of  the  same  degree  as  the  index  of 
the  required  root. 

1.  Kequired  the  cube  root  of  a^6V. 
We  have,  (abH^y  =  a^b^c^. 
Then,  by  §  206,  \/a^b^  =  ab^c\ 

2.  Required  the  fifth  root  of  -32a^ 
We  have,  (-2a)5  =  -32a^ 
Whence,                           \/-  32  a^  =  -  2  <?. 


EVOLUTION  191 

3.   Eequired  the  fourth  root  of  a^. 

We  have  either       (+  «)*  or  (-  ay  equal  to  a*. 

Whence,  Vo^  =  ±a. 

The  sign  ± ,  called  the  double  sign,  is  prefixed  to  an  expres- 
sion when  we  wish  to  indicate  that  it  is  either  +  or  — . 

209.   From  §  208,  we  have  the  following  rule : 

Extract  the  required  root  of  the  absolute  value  of  the  numerical 
coefficient,  and  divide  the  exponent  of  each  letter  by  the  index  of 
the  required  root. 

Give  to  every  even  root  of  a  positive  term  the  sign  ± ,  and  to 
every  odd  root  of  any  term  the  sign  of  the  term  itself. 

1.   Find  the  square  root  of  9  a''6V®. 


By  the  rule,  V9¥¥c^  =  ±  3 a^b^c^. 

2.   Find  the  cube  root  of  —64  x^y^. 


V—  64  x^y^"*  =  —  4  x^y^. 

The  root  of  a  large  number  may  sometimes  be  found  by 
resolving  it  into  its  prime  factors. 

3.   Find  the  square  root  of  254016. 


We  have,       V254016  =  VWxWxT^  =±2^xS^x7  =±  504. 


4.   Find  the  value  of  V72  X  75  X  135. 

y/12  X  76  X  136  =  \^(28  x  32)  x  (3  x  52)  x  (33  x  5) 


=  V28x  36  X  53  =  2  X  32  X  5  =  90. 

EXERCISE  84 

Find  the  values  of  the  following : 

1.  V36^.  4.    a/81  n%^y .  7.    v'64  a'7i'\ 

2.  a/64  a''b'c\  5.    Vl21  a'^b^'c*.  8.    -V^^^^2i3W. 


3.    -sZ-xy^z''.  6.    ^-216a^yz''.       9.    VlQ9x^y' 


192  ALGEBRA 

10.    A/'128  m%2^  13.    V2916.  16.    VSlx  64x324. 


11.  V343a^+^2/^.        14.    V30625.  17.    V84x  54x126. 

12.  ^625  a^'^b*^         15.    V86436.  18.    ^/5S32, 


19.    Vl5  xy  X  33  2/2!  X  55  zx. 

20.  -^21952.  23.    </l04976. 

21.  V627264.  24.    a/59049. 

22.  ■v/112  X  168  X  252.  25.    ■v/135  x  375  x  625. 


26.    V(a'  -  5  a  4-  6)(a^  +  2  a  -  8)(a2  +  a  - 12). 

210.   Any  Root  of  a  Fraction. 

It  follows  from  §  203  that,  to  find  any  root  of  a  fraction, 
each  of  whose  terms  is  a  perfect  power  of  the  same  degree  as 
the  index  of  the  required  root,  extract  the  required  root  of  both 
numerator  and  denominator. 


Ex. 


27  a^b'         V27a^         3  ab^ 


64  c^  ^64?  4c3 


EXERCISE  85 

Find  the  values  of  the  following: 

1. 


/64^  «      5/32  g^  K      V 


m" 


49  2*  >*  b'<^  \256n« 


"  "V     125  6«*  *  ^         2/''    *  '  ^'729  6«"* 

211.   We  have  Vipiy'  =  Va^  =  a*"  =  (-v/a«)"». 


^a;.     Eequired  the  value  of  V  (32  a^y. 

We  have,       v^(32  aio)4  =  ( ^^^32^*  =  (2  a2)4  =  16  a^. 

This  method  of  finding  the  root  is  shorter  than  raising 
32  a^^  to  the  fourth  power,  and  then  taking  the  fifth  root  of 
the  result. 


EVOLUTION  193 

EXERCISE  86 

Find  the  values  of  the  following : 

1.  </(<i4ar.         ^    ^(-243aVcy.      6.  ^'/f- l^Y. 

2.  V(I^)'.  ^^ ^^     27  W 

3.  </JW^¥f-      ^'  ^(^'^  "*""")'•  7.  V(a^-2a6  +  6»)l 

SQUARE  ROOT  OF  A  POLYNOMIAL 

212.   In  §  112,  we  showed  how  to  find  the  square  root  of  a 
trinomial  perfect  square. 

The  square  roots  of  certain  polynomials  of  the  form. 

a" -\-b'  -h c"  +  2 ab  +  2  ac4'2  be 

can  be  found  by  inspection. 

Ex.     Find  the  square  root  of 

9x^-\-y^-^4.z^-\-6xy  — 12  xz  — 4:  yz. 

"We  can  write  the  expression  as  follows  : 

j    (3x)2  +  ?/2+(_20)2  +  2(3x>  +  2(3x)(-2;?)  +22/(-20). 

By  §  204,  this  is  the  square  of  Sx -{■  y  +(— 2  z). 

Then,  the  square  root  of  the  expression  is  Sx  +  y  —  2  z. 

(The  result  could  also  have  been  obtained  in  the  form  2z  —  y  — 3x.) 

r^  EXERCISE  87 

Find  the  square  roots  of  the  following: 

1.  a'-{-b'-\-c'-2ab-2ac  +  2bc. 

2.  a^  +  4:y^-\-9  +  4:Xy-\-6x-\-12y. 

3.  l+25m2  +  36n2-10m  +  12w-60m7i. 

4.  a'  +  Slb'-\-16-{-lSab-Sa-72b, 

5.  9x^-{-y^-i-25z^-6xy-S0xz  +  10yz. 

6.  36  m^  +  64  n^  +  cc^  +  96  7nn  — 12  mx  — 16  nx, 

7.  16  a^  +  9  6^  +  81  c"  +  24  a'b''  +  72  aV  -f  54  ft^c^. 

8.  25a;«  +  49y«  +  36;2«-7Ox3/H-6Oa^3^-84  2/^0^ 


194  ALGEBRA 

213.  Square  Root  of  any  Polynomial  Perfect  Square. 

By  §  204,  (a  +  &  +  c)2  =  a-  +  2a&  +  62_|-  2ac  4-  2  6c  4-  c^ 

=  a2  +  (2a  +  6)6  +  (2a  +  26  +  c)c.    (1) 

Then,  if  the  square  of  a  trinomial  be  arranged  in  order  of 
powers  of  some  letter  : 

I.  The  square  root  of  the  first  term  gives  the  first  term  of 
the  root,  a. 

II.  If  from  (1)  we  subtract  a^,  we  have 

(2a  +  b)b-\-(2a-\-2b-\-c)c.  (2) 

The  first  term  of  this,  when  expanded,  is  2 ah-,  if  this  be 
divided  by  twice  the  first  term  of  the  root,  2  a,  we  have  the 
next  term  of  the  root,  b. 

III.  If  from  (2)  we  subtract  (2a  +  b)  b,  we  have 

(2a  +  2  6  +  c)c.  (3) 

The  first  term  of  this,  when  expanded,  is  2ac;  if  this  be 
divided  by  twice  the  first  term  of  the  root,  2  a,  we  have  the  last 
term  of  the  root,  c. 

IV.  If  from  (3)  we  subtract  (2  a -\- 2  b -{- c)  c,  there  is  no 
remainder. 

Similar  considerations  hold  with  respect  to  the  square  of  a 
polynomial  of  any  number  of  terms. 

214.  The^  principles  of  §  213  may  be  used  to  find  the  square 
root  of  a  polynomial  perfect  square  of  any  number  of  terms. 

Let  it  be  required  to  find  the  square  root  of 
4tx'  +  12x^-7i^-24.x-^16. 
4:x'-\-12x^-   7a^-24a;  +  16  |2ar^+3a;-4 


2a-^b  =  4.x^-{-Sx 
3x 


12a^-   7x2-24.15  +  16,  1st  Eem. 

12a^4-    9.^2 


2a  +  26  +  c  =  4ic2-|-6a5-4 
-4 


-  16  or  -  24  a;  +  16,  2d  Rem. 
-16ar'-24ic  +  16 


EVOLUTION  195 

The  first  term  of  the  root  is  the  square  root  of  4  ic*,  or  2  x"^. 

Subtracting  the  square  of  2  x^,  4  x*,  from  the  given  expression,  the  first 
remainder  is  \2x^  —  1  x-^  —  24  x  +  16. 

Dividing  the  first  term  of  this  by  twice  the  first  term  of  the  root,  4  x^, 
we  have  the  next  term  of  the  root,  3  x  (§  213,  II). 

Adding  this  to  4  x^  gives  4  x^  +  3  x  ;  multiplying  the  result  by  3  x,  and 
subtracting  the  product,  12  x^  +  9  x^,  from  the  first  remainder,  gives  the 
second  remainder,  —  16  x^  —  24  x  +  16. 

Dividing  the  first  term  of  this  by  twice  the  first  term  of  the  root,  4  x^, 
we  have  the  last  term  of  the  root,  -  4  (§  213,  III). 

If  from  the  second  remainder  we  subtract  (4x2  +  6x  —  4)(— 4),  or 
—  16  x2  -  24  X  +  16,  there  is  no  remainder  ;  then,  2  x^  +  3  x  -  4  is  the 
required  root  (§  213,  IV). 

The  expressions  4  x^  and  4  x^  +  6  x  are  called  trial-divisors^  and 
4  x2  +  3  X  and  4  x^  +  6  x  —  4  complete  divisors. 

We  then  have  the  following  rule  for  extracting  the  square 
root  of  a  polynomial  perfect  square : 

Arrange  the  expression  according  to  the  powers  of  some  letter. 

Extract  the  square  root  of  the  first  terrti,  write  the  result  as  the 
first  term  of  the  root,  and  subtract  its  square  from  the  given 
expression,  arranging  the  remainder  in  the  same  order  of  powers 
as  the  given  expression. 

Divide  the  first  term  of  the  remainder  by  twice  the  first  term  of 
the  root,  and  add  the  quotient  to  the  part  of  the  root  already  found, 
and  also  to  the  trial-divisor. 

Multiply  the  complete  divisor  by  the  term  of  the  root  last  obtained, 
and  subtract  the  product  from  the  remainder. 

If  other  terms  remain,  proceed  as  before,  doubling  the  part  of 
the  root  already  found  for  the  next  trial-divisor. 

215.   Examples. 

1.   Find  the  square  root  of  9  a;^  +  30  aV  +  25  a\ 

9  x*  +  30  a^x^  +  25  a^  |  3  x^  +  5  g^ 
•  9x4 


6x2  4-5a3 


30  a3x2 

30  a3cc2  +  25  a^ 


It  is  usual,  in  practice,  to  omit  those  terms,  after  the  first,  in  each 
remainder,  which  are  merely  repetitions  of  the  terms  in  the  given  expres- 
sion ;  thus,  in  the  first  remainder  of  Ex.  1,  we  leave  out  the  term  25  a®. 


196 


ALGEBRA 


It  is  also  usual  to  leave  out  of  the  written  work  the  multiplier  of  the 
complete  divisor. 

f^^     2.    Find  the  square  root  of 

Arranging  according  to  the  descending  powers  of  x,  we  have 

9  x6  -  12  a;5  +  28  a;*  -  22  a:3  +  20 ic2  -  8  ic  +  1  |  3x^- 2x2  + 4a;- 1 
9x6 


6x3-2x2 


-12x5 

-  12  x5  +    4  X* 


6  x3  -  4  x2  +  4  X 


24  X* 
24  X* 


16  x3  +  16  a;2 


6  x3  -  4  x2  +  8  X  -  1 


-  6x3+    4xz 

-  6x3+    4x2 


8x  +  l 


It  will  be  observed  that  each  trial-divisor  is  equal  to  the  pre- 
ceding complete  divisor  ivith  its  last  term  doubled. 
If,  in  Ex.  2,  we  had  written  the  expression 

1  -  8  X  +  20  x2  -  22  x3  +  28  x*  -  12  x6  +  9  x8, 

the  square  root  would  have  been  obtained  in  the  form  1— 4x  +  2x2  —  Sx^, 
which  is  the  negative  of  3  x3  —  2  x2  +  4  x  —  1. 


EXERCISE  88 
Find  the  square  roots  of  the  following : 

1.  x^-\-6a^-{-llx'-\-6x-\-l. 

2.  I_4a  +  2a^  +  4a3  +  a^ 

3.  9n^  +  12n«-20n^-16n  +  16. 

4.  5Qx'  +  4.-60x^-24:X-j-25x\ 

5.  x^  -\-  y^  -\-  4:Z^  —  2  xy  -}-  4:xz  —  4:yz. 

6.  Sa^-4.a-16a'  +  l-\-16a'-\-4.a^ 

7.  25x^  +  10a^y-hxy-\-30x^y^-\-6xY-^9f. 

8.  36a2_f.25  62  +  l6c2-60a6-48ac  +  406c. 

9.  9x*  +  6a^y-4t7a^y'^-16xf-i-64:y\ 

10.  127^-42n3  +  4_l9^^2  +  49^4 

11.  16a'  +  4:Sa^b  +  60a'b'  -\-36ab'  -h9b\ 


EVOLUTION  197 

12.  4  n''  -  16  nV  -f  36  n'^x^  -  40  n^x^  +  25  ar^. 

13.  30  xy  -  24  a;y  _  31  ^y  _}_  25  «» + 16  /. 

14.  4a;2  +  20a;  +  29  +  — +  i. 

X       x^ 

15.  a^_2a^-a*4-6a'^-3a2-4a  +  4. 

16.  5x^-23x'-{-12x-\-8a^-22o(^-\-16x^  +  4L, 

17  .2     2a5  ,  136^      46«  ,  46^ 

17.  a         ^4--^         9^  +  9^* 

18  ^'     ^'     41n^      5n     25 

■    4       3        36         6       16* 

19.    9  a«  +  6  a'x  +  31  aV  -  14  a^x^  + 17  aV  -  40  aa^  4- 16  a^. 

*    16     4.y'^20y^     5f     25y'' 

21     25     15a     41a^     3a'  ^     a' 
'4         b         4.b^      2^^     16b^' 

22.   44  a^^^  +  4  6«  -  30  a^ft  +  4  a^d^  +  25  a^  - 16  a6^  -  31  a*b\ 


SQUARE  ROOT  OF  AN  ARITHMETICAL  NUMBER 

216.  The  square  root  of  100  is  10 ;  of  10000  is  100 ;  etc. 
Hence,  the  square  root  of  a  number  between  1  and  100  is 

between  1  and  10;  the  square  root  of  a  number  between  100 
and  10000  is  between  10  and  100;  etc. 

That  is,  the  integral  part  of  the  square  root  of  an  integer  of 
one  or  two  digits,  contains  one  digit ;  of  an  integer  of  three  or 
four  digits,  contains  two  digits ;  and  so  on. 

Hence,  if  a  point  be  placed  over  every  second  digit  of  an  integer, 
beginning  at  the  units^  place,  the  number  of  points  shows  the  number 
of  digits  in  the  integral  part  of  its  square  root. 

217.  Square  Root  of  any  Integral  Perfect  Square. 

The  square  root  of  an  integral  perfect  square  may  be  found 
in  the  same  way  as  the  square  root  of  a  polynomial. 
Required  the  square  root  of  106929. 


198 


ALGEBRA 


2  a  +  6  =       600  +  20 
20 


106929 
a2  =  90000 


2a+26  +  c  =  600  +  40  +  7 

7 


16929 
12400 


300  +  20  +  7 
=  a-{'b-\-c 


4529 
4529 


Pointing  the  number  in  accordance  with  the  rule  of  §  216,  we 
find  that  there  are  three  digits  in  its  square  root. 

Let  a  represent  the  hundreds'  digit  of  the  root,  with  two 
ciphers  annexed;  b  the  tens'  digit,  with  one  cipher  annexed; 
and  c  the  units'  digit. 

Then,  a  must  be  the  greatest  multiple  of  100  whose  square  is 
less  than  106929 ;  this  we  find  to  be  300. 

Subtracting  a^,  or  90000,  from  the  given  number,  the  result 
is  16929. 

Dividing  this  remainder  by  2  a,  or  600,  we  have  the  quotient 
28+  ;  which  suggests  that  b  equals  20. 

Adding  this  to  2  a,  or  600,  and  multiplying  the  result  by  b,  or 
20,  we  have  12400 ;  which,  subtracted  from  16929,  leaves  4529. 

Since  this  remainder  equals  (2a  +  2  6  +  c)c  (§  213,  III),  we 
can  get  c  approximately  by  dividing  it  by  2  a  +  2  6,  or  600  +  40. 

Dividing  4529  by  640,  we  have  the  quotient  7+;  which 
suggests  that  c  equals  7. 

Adding  this  to  600  +  40,  multiplying  the  result  by  7,  and 
subtracting  the  product,  4529,  there  is  no  remainder. 

Then,  300  +  20  +  7,  or  327,  is  the  required  square  root. 

218.  Omitting  the  ciphers  for  the  sake  of  brevity,  and  con- 
densing the  operation,  we  may  arrange  the  work  of  the  example 
of  §  217  as  follows : 

106929  [327 
9 


62 

169 
124 

647 

4529 
4529 

EVOLUTION  199 

The  numbers  600  and  640  are  called  trial-divisors^  and  the  numbers 
620  and  647  are  called  complete  divisors. 

We  then  have  the  following  rule  for  finding  the  square  root 
of  an  integral  perfect  square : 

Separate  the  number  into  periods  by  pointing  every  second  digits 
beginning  with  the  units'  place. 

Find  the  greatest  square  in  the  left-hand  period,  and  write  its 
square  root  as  the  first  digit  of  the  root;  subtract  the  square  of  the 
first  root-digit  from  the  left-hand  period,  and  to  the  result  annex 
the  next  period. 

Divide  this  remainder,  omitting  the  last  digit,  by  twice  the  part 
of  the  root  already  found,  and  annex  the  quotient  to  the  root,  and 
also  to  the  trial-divisor. 

Multiply  the  complete  divisor  by  the  root-digit  last  obtained,  and 
subtract  the  product  from  the  remainder. 

If  other  periods  remain,  proceed  as  before,  doubling  the  part  of 
the  root  already  found  for  the  next  trial-divisor. 

Note  1.  It  sometimes  happens  that,  on  multiplying  a  complete  divisor 
by  the  digit  of  the  root  last  obtained,  the  product  is  greater  than  the 
remainder. 

In  such  a  case,  the  digit  of  the  root  last  obtained  is  too  great,  and  one 
less  must  be  substituted  for  it. 

Note  2.  If  any  root-digit  is  0,  annex  0  to  the  trial-divisor,  and  annex 
to  the  remainder  the  next  period.     (See  the  illustrative  example  of  §  220.) 

219.  Ex.     Find  the  square  root  of  4624. 

4624  L68 
36 


128 


1024 
1024 


The  greatest  square  in  the  left-hand  period  is  36. 

Then  the  first  digit  of  the  root  is  6. 

Subtracting  62,  or  36,  from  the  left-hand  period,  the  result  is  10 ;  to 
this  we  annex  the  next  period,  24. 

Dividing  this  remainder,  omitting  the  last  digit,  or  102,  by  tv^^ice  the 
part  of  the  root  already  found,  or  12,  the  quotient  is  8  ;  this  we  annex  to 
the  root,  and  also  to  the  trial-divisor. 


200  ALGEBRA 

Multiplying  the  complete  divisor,  128,  by  8,  and  subtracting  the  product 
from  the  remainder,  there  is  no  remainder. 
Then,  68  is  the  required  square  root. 

220.   We  will  now  show  how  to  find  the  square  root  of  a 
number  which  is  not  integral. 

Ex.     Find  the  square  root  of  49.449024. 


We  have,         ^4-9:149024  =  J4M49024  ^  V49449024. 
V  1000000       vioooooo 


49449024  |  7032 
49 


1403 

4490 
4209 

14062 

28124 
28124 

Since  14  is  not  contained  in  4,  we  write  0  as  the  second  root-digit,  in 
the  above  example  ;  we  then  annex  0  to  the  trial-divisor  14,  and  annex  to 
the  remainder  the  next  period,  90.    (See  Note  2,  §  218.) 

Then, 


'49.449024  =  """"  =  7.032 
1000 

mged  as  follows : 

49.449024  |  7.032 

49 

1403 

4490 

4209 

14062 

28124 

28124 

Then,  if  a  point  he  placed  over  every  second  digit  of  miy 
number,  beginning  with  the  units'  place,  and  extending  in  either 
direction,  the  rule  of  §  218  may  be  applied  to  the  result  and  the 
decimal  point  inserted  in  its  proper  position  in  the  root. 

^^  EXERCISE  89 

Find  the  square  roots  of  the  following : 
1.   5776.  2.   15376.  3.   67081. 


EVOLUTION 

201 

4. 

8427.24. 

8. 

7974.49. 

12. 

.30316036. 

5. 

.165649. 

9. 

.00459684. 

13. 

39.375625. 

6. 

.133225. 

10. 

22014864. 

14. 

.000064272289. 

7. 

54.4644. 

11. 

1488.4164. 

15. 

889060.41. 

221.   Approximate  Square  Roots. 

If  there  is  a  final  remainder,  the  number  has  no  exact  square 
root ;  but  we  may  continue  the  operation  by  annexing  periods 
of  ciphers,  and  obtain  an  approximate  root,  correct  to  any  desired 
number  of  decimal  places. 

Ex.    Find  the  square  root  of  12  to  four  decimal  places. 

3.4641  + 


12. 
9 

00000000 

64 

3  00 
2  56 

686 

4400 
4116 

6924 

28400 
27696 

69281  I  70400 

222.  The  approximate  square  root  of  a  fraction  may  be  ob- 
tained by  taking  the  square  root  of  the  numerator,  and  then 
of  the  denominator,  and  dividing  the  first  result  by  the  second. 

If  the  denominator  is  not  a  perfect  square,  it  is  better  to 
reduce  the  fraction  to  an  equivalent  fraction  whose  denominator 

I  is  a  perfect  square. 
Ex.     Find  the  value  of  Vf  to  five  decimal  places. 
; 


Vie 

EXERCISE  90 
Find  the  first  five  figures  of  the  square  root  of : 

1.  2.  4.   17.  7.   .3.  10.   .008. 

2.  5.  5.   59.  8.    .067.  11.   .00095. 

3.  11.  6.   75.8.  9.   .46.  12.  96.756. 


202  ALGEBRA 

Find  the  first  four  figures  of  the  square  root  of : 

13.  i.  15.  II-  17.  f.  19.  ii.  21.  If 

14.  |.  16.  1  18.  |.  20.  i|.  22.  if. 

CUBE   ROOT  OF  A  POLYNOMIAL 

223.  The  cube  roots  of  certain  polynomials  of  the  form 

a^  +  Sa'b-\-3ab^  +  b^ 
can  be  found  by  inspection. 

Ex.     Find  the  cube  root  of  8  a^  -  36  a^b^  +  54  a&*  -  27  b\ 
We  can  write  the  expression  as  follows  : 

(2  ay  -  3  (2  a)2(3  62)  +  3  (2  a)  (3  62)2  _  (3  52)8. 

By  §  205,  this  is  the  cube  of  2  a  -  3  62. 

Then,  the  cube  root  of  the  expression  is  2  a  —  3  62. 

EXERCISE  91 
Find  the  cube  roots  of  the  following  : 

1.  a3  +  6a2  +  12a  +  8. 

2.  l-9m-^27m'-27m^ 

3.  647i3-48n2  +  12n-l. 

4.  125a^  +  75afy  +  15xy^  +  f. 

5.  a«  +  18a*6^4-108a26«4-216  6». 

6.  125m3+150m2w  +  60mn2  +  87i3. 

7.  27  a^b^-lOSa'b'c  +  lUabc'- 64.  (^, 

8.  m«-21mV  +  147mV-343a;^2 

224.  Cube  Root  of  any  Polynomial  Perfect  Cube. 

By§205,  (a  +  6  +  c)«=[(a  +  &)  +  c7 
=  (a  +  6)3  +  3(a  +  6)2c  +  3(a  +  6)c2H-c3 
=  a»  +  3a26  +  3a62  +  &'  +  3(a4-&)'c  +  3(a-f&)c'  +  c' 
=  a^  +  (3  a2  +  3  a6  +  6^)6  +  [3(a  +  bf  +  3(a  +  &)c  +  c^Jc  (1) 


EVOLUTION  203 

Then,  if  the  cube  of  a  trinomial  be  arranged  in  order  of 
powers  of  some  letter : 

I.  The  cube  root  of  the  first  term  gives  the  first  term  of  the 
cube  root,  a. 

II.  If  from  (1)  we  subtract  a^,  we  have 

(3  a^  +  3  a&  +  b^)b  [f  [3(a  -f  bf  +  3(a  +  b)c  -f  c'y.        (2) 

The  first  term  of  this,  "^when  expanded,  is  3  a^b ;  if  this  be 
divided  by  three  times  the  square  of  the  first  term  of  the  root, 
3  a^,  we  have  the  next  term  of  the  root,  b. 

III.  If  from  (2)  we  subtract  (3  a^  +  3  a5  +  b^b,  we  have 

[3(a  +  6)2  +  3(a  +  b)c  +  c']c.  (3) 

The  first  term  of  this,  when  expanded,  is  3a^c;  if  this  be 
divided  by  three  times  the  square  of  the  first  term  of  the  root, 
3  a^,  we  have  the  last  term  of  the  root,  c. 

IV.  If  from  (3)  we  subtract  [3(a  +  6)^  +  3(a  +  6)c  +  c2]c, 
there  is  no  remainder. 

Similar  considerations  hold  with  respect  to  the  cube  of  poly- 
nomial of  any  number  of  terms. 

225.  The  principles  of  §  224  may  be  used  to  find  the  cube 
root  of  a  polynomial  perfect  cube  of  any  number  of  terms. 
Let  it  be  required  to  find  the  cube  root  of 

x^+6  #+  3  X*  -  28  x8  -  9  x2 + 54  x-  27 


3  a2+3  a6  +  62=  3  x*+6  x3+4  x^ 

'  2x 


6a:5+  3cc4-28x3-9ic2+64x-27 
6x5  +  12x4+  8x3  


3(a+ 6)2=3  x*+12  x3+12  x2 
3(a  +  6)c+c2= -  9x2-18x+9 


3x4+12x8+  3x3-18x+9 
-3 


9  x4-36  x8-9  x2+54  x-27^ 
9  x4-36  x8-9  x2+54  x~27 


The  first  term  of  the  root  is  the  cube  root  of  x^,  or  x'^. 
Subtracting  the  cube  of  x^,  or  x^,  from  the  given  expression,  the  first 
jmainder  is  6  x&  +  3  x*  -  28  xs  -  9  x'^  +  64  x  -  27. 


204  ALGEBKA 

Dividing  the  first  term  of  this  by  three  times  the  square  of  the  first 
term  of  the  root,  3  x'^,  we  have  the  next  term  of  the  root,  2  x  (§  224,  II). 

Now,  3  a6  +  62  equals  3  x  x^  x  2  ic  +  (2  x)2,  or  6  x^  +  4  x^. 

Adding  this  to  3  x^,  multiplying  the  result  by  2  x,  and  subtracting  the 
product,  6  x^  +  12  X*  +  8  x^,  from  the  first  remainder,  gives  the  second 
remainder,  -  9  x*  -  36  x^  -  9  x2  +  54  x  -  27  (§  224,  III). 

Dividing  the  first  term  of  this  by  three  times  the  square  of  the  first 
term  of  the  root,  3x2,  we  have  the  last  term  of  the  root,   —3. 

Now,  3(a+6)2  equals  3(x2  +  2x)2,  or  3  x*  + 12x3+ 12x2;  3(a  +  6)c 
equals  3(x2  +  2  x) ( -  3),  or  -  9  x2  -  18  x  ;  and  d^  =  9. 

Adding  these  results,  we  have  3  x*  +  12  x^  +  3  x2  —  18  x  +  9. 

Subtracting  from  the  second  remainder  the  product  of  this  by  —  3,  or 
—  9  X*  —  36  x^  —  9  x'-^  +  54  X  —  27,  there  is  no  remainder ;  then,  x2  +  2  x  —  3 
is  the  required  root  (§  224,  IV). 

The  expressions  3  x*  and  3  x*  +  12  x'  +  12  x2  are  called  trial-divisors, 
and  the  expressions  3  x*  +  6  x^  +  4  x2  and  3  x*  +  12  x^  +  3  x2  -  18  x  +  9 
complete  divisors. 

We  then  have  the  following  rule  for  finding  the  cube  root  of 
a  polynomial  perfect  cube : 

Arrange  the  expression  according  to  the  powers  of  some  letter. 

Extract  the  cube  root  of  the  first  term,  write  the  result  as  the 
first  term  of  the  root,  and  subtract  its  cube  from  the  given 
expression;  arranging  the  remainder  in  the  same  order  of  powers 
as  the  given  expression. 

Divide  the  first  teryn  of  the  remainder  by  three  times  the  square 
of  the  first  term  of  the  root,  and  write  the  result  as  the  next  term 
of  the  root. 

Add  to  the  trial-divisor  three  times  the  product  of  the  term  of 
the  root  last  obtained  by  the  part  of  the  root  previously  found,  and 
the  square  of  the  term  of  the  root  last  obtained. 

Multiply  the  complete  divisor  by  the  term  of  the  root  last 
obtained,  and  subtract  the  product  from  the  remainder. 

If  other  terms  remain,  proceed  as  before,  taking  three  times 
the  square  of  the  part  of  the  root  already  found  for  the  next  trial- 
divisor. 

226.   Examples. 

1.   Find  the  cube  root  of  8  a;^  -  36  xhj  +  54  a^y  -  27  f. 


EVOLUTION 


205 


8  x6  -  36  x*^  +  54  a;V  _  27  y^  \2x^-Sy 
8x6 


12  X*  -  18  x2?/  +  9  2/2 


-  36  x% 

-  36  x^y  +  54  x^y"^  -  27  yS 


It  is  usual,  in  practice,  to  omit  those  terms,  after  the  first,  in  each 
remainder,  which  are  merely  repetitions  of  the  terms  in  the  given  expres- 
sion ;  and  also  to  leave  out  of  the  written  work  the  multiplier  of  the  com- 
plete divisor. 

2.   Find  the  cube  root  of  40  aj^  -  6  or'  -  64  +  «« -  96  a;. 
Arranging  according  to  the  descending  powers  of  x,  we  have 


x6-6x 

X6 

>  +  40  x3  -  96  X  -  64  1  x2  -  2  X  - 

5 

'  -f  12  x*  -  8  x8 

^ 

3  x4  -  6  x3  +  4  x2 

-6x 
-6x 

3x4- 

-  12  x3  +  12  x2 

-  12  x2  +  24  X  +  16 

-  12  x*  -f  48  x8 

-  12  x*  -F  48  x3  -  96  X  -  64 

3x4- 

-12x3              -l-24a 

,+  16 

EXERCISE  92 

Find  the  cube  roots  of  the  following : 
1.   27a;«-f27a;^-h9aj2-fl.       2.  S  a' - 60 a'b +150 a'b'' -125  b\ 

3.  S36xy'^3^3a:^-64:f-5SSx'y. 

4.  x^  +  9a^-\-30x'  +  4.5a^  +  S0a:^-\-9x-\-l. 

5.  Sn^-12n'-30n''  +  357i^  +  4.5n^-2Tn-27. 

6.  9a'  +  54.a'-l-2Sa^-3d'+27a'  +  6a, 


7. 


8. 


27     12 


aJl      b^ 
16      64* 


n^  - 12  n'x  +  57  nV  - 136  n^a^  + 171  nV-108  nar'+27  a^. 
9.   135  a%^  -f- 12  a'b  - 125  b^  +  Sa^-59  a^b^+75  ab'-54.  a%K 
10.   152a^-27-63aj2  4.27a;6-h63a5^-108a;-108a;^. 


ill.    ^- 


15  a" 


153a^^()     153_15      1. 


8         4      ■      4      '       '   '   2a      a'      W 

12.  64mH144m^-h204m*-hl71m3-h92m2-}-36m-h8. 

13.  a;9-6a;«-M5a;^-26a;«+39a^-42a;^-}-37a^-30a^-f-12aj- 


206 


ALGEBRA 


CUBE  ROOT   OF  AN  ARITHMETICAL  NUMBER 

227.  The  cube  root  of  1000  is  10;  of  1000000  is  100;  etc. 
Hence,  the  cube  root  of  a  number  between  1  and  1000  is 

between  1  and  10;  the  cube- root  of  a  number  between  1000 
and  1000000  is  between  10  and  100 ;  etc. 

That  is,  the  integral  part  of  the  cube  root  of  an  integer  of 
one,  two,  or  three  digits,  contains  one  digit ;  of  an  integer  of 
four,  five,  or  six  digits,  contains  two  digits ;  and  so  on. 

Hence,  if  a  point  he  placed  over  every  third  digit  of  an  integer ^ 
beginning  at  the  units''  place,  the  number  of  points  shows  the  number 
of  digits  in  the  integral  part  of  its  cube  root. 

228.  Cube  Root  of  any  Integral  Perfect  Cube. 

The  cube  root  of  an  integral  perfect  cube  may  be  found  in 
the  same  way  as  the  cube  root  of  a  polynomial. 
Eequired  the  cube  root  of  12487168. 

200  +  30  +  2 
=  a  +  &  +  c 


12487168 
a^  =  8000000 

3^2  =  120000 

Sab=    18000 

¥  =   900 

4487168 

138900 
30 

4167000 

3  (a +  6)2=  158700 

3(a  +  6)c=   1380 

c2=      4 

320168 

160084 
2 

320168 

Pointing  the  number  in  accordance  with  the  rule  of  §  227,  we 
find  that  there  are  three  digits  in  the  cube  root. 

Let  a  represent  the  hundreds'  digit  of  the  root,  with  two 
ciphers  annexed ;  b  the  tens'  digit,  with  one  cipher  annexed  ; 
and  c  the  units'  digit. 

Then,  a  must  be  the  greatest  multiple  of  100  whose  cube  is 
less  than  12487168 :  this  we  find  to  be  200. 


EVOLUTION  207 

Subtracting  a^,  or  8000000,  from  the  given  number,  the  result 
is  4487168. 

Dividing  this  by  3  a^,  or  120000,  we  have  the  quotient  37+  ; 
which  suggests  that  h  equals  30. 

Adding  to  the  divisor  120000,  3  ah,  or  18000,  and  h\  or  900, 
we  have  138900. 

Multiplying  this  by  6,  or  30,  and  subtracting  the  product 
4167000  from  4487168,  we  have  320168. 

Since  this  remainder  equals  [3(a4- 6)^  +  3  (a -f  6)  c  +  c^]c 
(§  224,  III),  we  can  get  c  approximately  by  dividing  it  by 
3  (a +  6)2,  or  158700. 

Dividing  320168  by  158700,  the  quotient  is  2+  ;  which  sug- 
gests that  c  equals  2. 

Adding  to  the  divisor  158700,  3(a  +  6)c,  or  1380,  and  c^,  or 
4,  we  have  160084 ;  multiplying  this  by  2,  and  subtracting  the 
product,  320168,  there  is  no  remainder. 

Then,  200  +  30  +  2,  or  232,  is  the  required  cube  root. 

229.  Omitting  the  ciphers  for  the  sake  of  brevity,  and  con- 
densing the  process,  the  work  of  the  example  of  §  228  will 
stand  as  follows : 

12487168  [232 
8 


1200 

4487 

180 

9 

1389 

4167 

15870 

0 

320168 

1380 

4 

16008 

4 

320168 

The  numbers  120000  and  168700  are  called  trial-divisors,   and  the 
numbers  138900  and  160084  are  called  complete  divisors. 

We  then  have  the  following  rule  for  finding  the  cube  root  of 
an  integral  perfect  cube : 

Separate  the  number  inta  periods  by  pointing  every  third  digit, 
beginning  with  the  units'  place. 


208  ALGEBRA 

Find  the  greatest  cube  in  the  left-hand  2)eriod,  and  write  its  cube 
root  as  the  first  digit  of  the  root;  subtract  the  cube  of  the  first  root- 
digit  from  the  left-hand  period,  and  to  the  result  annex  the  next 
period. 

Divide  this  remainder  by  three  times  the  square  of  the  part  of 
the  root  already  found,  ivith  ttvo  ciphers  annexed,  and  write  the 
quotient  as  the  next  digit  of  the  root. 

Add  to  the  trial-divisor  three  times  the  product  of  the  last  root- 
digit  by  the  part  of  the  root  previously  found,  with  one  cipher 
annexed,  and  the  square  of  the  last  root-digit. 

Multiply  the  complete  divisor  by  the  digit  of  the  root  last 
obtained,  aiid  subtract  the  product  from  the  remainder. 

If  other  periods  remain,  proceed  as  before,  taking  three  times 
the  square  of  the  part  of  the  root  already  found,  with  two  ciphers 
annexed,  for  the  next  trial-divisor. 

Note  1.    Note  1,  §  218,  applies  with  equal  force  to  the  above  rule. 

Note  2.  If  any  root-figure  is  0,  annex  two  ciphers  to  the  trial- 
divisor,  and  annex  to  the  remainder  the  next  period. 

230.  In  the  example  of  §  228,  the  first  complete  divisor  is 

The  next  trial-divisor  is  3  (a  4-  b^,  or  3  a^  +  6  a&  +  3  b^. 

This  may  be  obtained  from  (1)  by  adding  to  it  its  second 
term,  and  double  its  third  term. 

That  is,  if  the  first  number  and  the  double  of  the  second  number 
required  to  complete  any  trial-divisor  be  added  to  the  complete 
divisor,  the  result,  with  two  ciphers  annexed,  will  give  the  next 
trial-divisor. 

This  rule  saves  much  labor  in  forming  the  trial-divisors. 

231.  Ex.     Find  the  cube  root  of  157464. 

157464  [54 
125 


7500 

32464 

600 

16 

8116 

32464 

EVOLUTION 


209 


232.   We  will  now  show  how  to  find  the  cube  root  of   a 
number  which  is  not  integral. 

Ex.     Find  the  cube  root  of  8144.865728. 


We  have,   v^8144. 865728  =  ^ 


8144865728      V8144865728 


1000000 


v^lOOOOOO 


8144865728  |  2012 
8 


120000 

600 

1 

120601 

144865 
120601 

600 
2 

24264728 

12120300 

12060 

4 

12132364 

24264728 

Since  1200  is  not  contained  in  144,  we  write  0  as  the  second  root-digit, 
in  the  above  example ;  we  then  annex  two  ciphers  to  the  trial-divisor 
1200,  and  annex  to  the  remainder  the  next  period,  865.     (Note  1,  §  229.) 

The  second  trial-divisor  is  formed  by  the  rule  of  §  230. 

Adding  to  the  complete  divisor  120601  the  first  number,  600,  and  twice 
the  second  number,  2,  required  to  complete  the  trial-divisor  120000,  we 
have  121203  ;  annexing  two  ciphers  to  this,  the  result  is  12120300. 

Then,  ^8144.865728  =  ?^  =  20.12. 

The  work  may  be  arranged  as  follows : 

18144. 865728  |  20.12 


8 

120000 

144  865 

600 

1 

120601 

120  601 

600 

24  264728 

2 

12120300 

12060 

4 

12132364 

24  264728 

210 


ALGEBRA 


It  follows  from  the  above  that,  if  a  point  he  placed  over  every 
third  digit  of  any  number,  beginning  with  the  units'*  place,  and 
extending  in  either  direction,  the  rule  of  §  229  may  be  applied  to 
the  result,  and  the  decimal  point  inserted  in  its  proper  position  in 
the  root. 

EXERCISE  93 

Find  the  cube  roots  of  the  following : 


1.   54872. 

6. 

3176523. 

11.  331373888. 

2.  262144. 

7. 

130323.843. 

12.   37.595375. 

3.   103.823. 

8. 

.102503232. 

13.   667627.624. 

4.   .884736. 

9. 

.000356400829. 

14.    .964430272. 

5.   .000493039 

10. 

22.665187. 

15.   3422470.843. 

Find  the  first  four  figures  of  the  cube  root  of : 

16.  4. 

18. 

8.2. 

20.    J. 

22.   A. 

17.  9. 

19. 

.03. 

21.  M. 

23.  |. 

233.  If  the  index  of  the  required  root  is  the  product  of  two 
or  more  numbers,  we  may  obtain  the  result  by  successive  ea>- 
tractions  of  the  simpler  roots. 

For  by  §206,  C\/^)'»"  =  a.  « 

Taking  the  nth  root  of  both  members,  j 

CVS)»=V^.  (1) 

Taking  the  mth  root  of  both  members  of  (1), 

Hence,  the  mnth  root  of  an  expression  is  equal  to  the  mth  root 
of  the  nth  root  of  the  expression. 

Thus,  to  find  the  fourth  root  of  an  expression,  we  find  the 
square  root  of  its  square  root ;  to  find  the  sixth  root,  we  find 
the  cube  root  of  the  square  root,  etc. 


EVOLUTION  211 

EXERCISE  94 

Find  the  fourth  roots  of  the  following : 

1 .  a' -16  a'W  +  96  a^h'  -  256  a%^  +  256  h^. 

2.  81  a«  - 108  a'  + 162  a« - 120  a'  +  91  a^  -  40  a^+lS  o?-^  a 
+  1. 

3.  16  +  32  a;  -  72  a^  - 136  aj3  + 145  a;^+204  ^-162  a;«-108  x' 

+  81a^. 

4.  .011156640625. 

Find  the  sixth  roots  of  the  following : 

6.   64a;i2_^192a;i«  +  240a^  +  160a^  +  60a;^4-12a;2  +  l- 

6.  a«  - 18  a^  + 135  a^  -  540  a^  + 1215  a?  - 1458  a  4-  729. 

7.  34296.447249. 

234.  By  §206,       (•v/^)"  =  a&. 
Also,  {-y/a  X  -y/ly  =  (-v/a)"  X  (V&)"  =  «&. 

Then,  (-v/^)«  =  ( v^  x  ■>/&)". 

Whence,  Voft  =  Va  x  ">/&. 


Y 

212  ALGEBRA 


XVII.    THEORY  OF  EXPONENTS 

235.  In  the  preceding  portions  of  the  work,  an  exponent  has 
been  considered  only  as  a  positive  integer. 

Thus,  if  m  is  a  positive  integer, 

a"*  =  a  X  a  X  a  X  •••  to  m  factors.  (§  11) 

The  following  results  have  been  proved  to  hold  for  any- 
positive  integral  values  of  m  and  n : 

orxor  =  «"*+"  (§  bOf).  (1) 

{w^y  =  or^  (§  93).  (2) 

236.  It  is  necessary  to  employ  exponents  which  are  not 
positive  integers;  and  we  now  proceed  to  define  them,  and 
prove  the  rules  for  their  use. 

In  determining  what  meanings  to  assign  to  the  new  forms,  it 
will  be  convenient  to  have  them  such  that  the  above  law  for 
multiplication  shall  hold  with  respect  to  them. 

We  shall  therefore  assume  equation  (1),  §  235,  to  hold  for 
all  values  of  m  and  n,  and  find  what  meanings  must  be  attached 
in  consequence  to  fractional^  negative,  and  zero  exponents. 

237.  Meaning  of  a  Fractional  Exponent. 

Let  it  be  required  to  find  the  meaning  of  a^. 

If  (1),  §  235,  is  to  hold  for  all  values  of  m  and  n, 

a^  X  a^  X  a^  =  a^^^"^^  =  a^. 

Then,  the  third  power  of  a^  equals  a^. 

Hence,  a^  must  be  the  cube  root  of  a^,  or  (3  =  V^. 

We  will  now  consider  the  general  case. 

p 
Let  it  be  required  to  find  the  meaning  of  a^,  where  p  and  q 

are  any  positive  integers. 


THEORY  OF   EXPONENTS  213 

If  (1),  §  235,  is  to  hold  for  all  values  of  m  and  w, 

?  P  P  ?+?+?+...  to  ?t.rm«  P-^q 

a^  X  a^  X  a'^  X  •••  to  g  factors  =  a*  '  *  =  a'     =  a^. 

I' 
Then,  the  gth  power  of  a*  equals  a^. 

p  p         

Hence,  a'  must  be  the  ^th  root  of  a^,  or  a'  —-{/a^. 

i.  Hence,  in   a  fractional   exponent,    the  numerator  denotes  a 
power,  and  the  denommator  a  root. 

For  example,  a^  =  -\/a^ ;  6^  =  V6^ ;  x^  =  -y/x ;  etc. 

EXERCISE  95 

Express  the  following  with  radical  signs : 

1.  aK       3.    7  mi       5.    ah^.       7.    8aW.  9.   x^yh^. 

2.  x^.       4.   5xk        6.   a^V'^      8.    lOw^cc^^.      10.   2a'"62cT. 

Express  the  following  with  fractional  exponents : 

11.  ^6.  13.    V^3.  15.   S</F.  17.   9^m^;?. 

12.  </a.  14.    ^n^  16.   4^/p.  18.    -^^"v//. 

19.    -^a->y6^«.  20.    ^s/^Vy^V^^. 

^^,   -  238.   Meaning  of  a  Zero  Exponent. 
i,(sA^<  If  (1),  §  235,  is  to  hold  for  all  values  of  m  and  n,  we  have 

a'^x  a^  =  a"^-^^  =  a*". 

Whence,  a»  =  — =  1.         ' 

a"* 

We  must  then  define  a^  as  being  equal  to  1. 

239.   Meaning  of  a  Negative  Exponent. 

et  it  be  required  to  find  the  meaning  of  a~^. 
If  (1),  §  235,  is  to  hold  for  all  values  of  m  and  n, 

a-^  X  a^  =  a-^+^  =  a«  =  1  (§  238). 


214  ALGEBRA 

1 


Whence, 


.-3 


y3 


Whence,  a~'-= 


a" 

We  will  now  consider  the  general  case. 

Let  it  be  required  to  find  the  meaning  of  a~%  where  s  repre- 
sents a  positive  integer  or  a  positive  fraction. 
If  (1),  §  235,  is  to  hold  for  all  values  of  m  and  n, 

a-'  xa'  =  a-'+'  =  a«  =  1  (§  238). 

1_^ 

a"' 

We  must  then  define  a~'  as  being  equal  to  1  divided  by  a'. 
For  example,  a"^  =  — ;  a  »=__;  333-1^"^  =  — _j  etc. 

240.   It  follows  from  §  239  that 

Any  f(xctor  of  the  numerator  of  a  fraction  may  be  transferred 
to  the  denominator,  or  any  factor  of  the  denominator  to  the 
numerator,  if  the  sign  of  its  exponent  be  changed. 

Thus,  a?l^J^^^^^^^  ,te. 

EXERCISE  96 
Express  with  positive  exponents : 


1. 

oj-y. 

5. 

a-^m-\ 

9. 

m~^n-^. 

2. 

ah-K 

6. 

m-iV. 

10. 

8a-^6-^V. 

3. 

m  *?l^ 

7. 

^a-^n-\ 

11. 

6  mT^nT^Q^. 

4. 

Sn-^a;. 

8. 

5  a; V^2;-^ 

12. 

la-^n-^x-^. 

Transfer  all  literal  factors  from  the  denominators  to  the 
numerators  in  the  following: 

13.  i.  14.  4-  15-  -^  16.  A- 


THEORY  OF  EXPONENTS  215 

17.  -^.         18.  -^^.         19.   ^-^.         20.   ^^"'^"l 

^      Transfer   all   literal   factors   from   the    mimerators    to   the 
denominators  in  the  following: 

21     i^         23    ^~^^  25     ^  a~^6^         07     9  m~^yi~^ 

22.  ^.  24.  ^.  26.  ^.  28.  A^^. 

241.  We  obtained  the  definitions  of  fractional,  zero,  and 
negative  exponents  by  supposing  equation  (1),  §  235,  to  hold 
for  such  exponents. 

Then,  for  any  values  of  m  and  n^ 

oTxa^'^z  or+\  (1) 

The  formal  proof  of  this  result  for  positive  or  negative,  integral  or 
fractional,  values  of  m  and  n  will  be  found  in  §  445. 

1.  Find  the  value  of  a^  x  a~^. 

We  have,  a^  x  a~^  =  a^-s  —  ^5-8, 

2.  Find  the  value  of  a  x  Va^. 

By  §237,  axVa^  =  axa^  =  a}-^^  =  a^, 

3.  Multiply  a  +  2  a*-  3  a^  by  2  -  4  a"^  -  6  a"^. 


4  a  ^-6  a 


2  a  +  4  a^  -    Qa^ 
-  4  a^  -    8  a^  +  12 


6  a^  -  12  +  18  a"^ 


2  a  -20a^  -f  18  a 


It  must  be  carefully  observed,  in  examples  like  the  above,  that  the 
zero  power  of  any  number  equals  1  (§  238). > 


216  ALGEBRA 

EXERCISE  97 

Multiply  the  following : 
1.   a^  by  a^.  ^'  4.   n^  by  n-^  7.   2x~^  hj  7  x\ 

5.  3  a-'  by  a'l  8.   x-^  by  -J/^. 

6.  m  by  4  m  ^".  9.    v^  by  a^. 
13.  3a?"%~^  by  4:X~^y. 

,,    '(/u.   a-^^/^  by  a-«V^. 
15.   m~^n~^^  by 


\f '• 

X-'  by  x-l 

4  3. 

a^  by  a~^. 

Vi 

\j    10. 

m^  by  — -• 

11. 

8Va-3  by 

12.  6  a^z,  by  a-^i'^.  3  nr'n^ 

16.  a;^  —  2  ic32/3  +  4  2/^  by  x^-\-2yK 

17.  2n-^-5-6nHy  3n-?-4. 

18.  4a-''  +  10a-2  +  25  by  2  a-' -5. 

19.  a^  -  ah^  4-  &"^  by  a^  +  ah^  +  6^ 

20.  x~^  —  x~^y^  +  2  x~^y^  by  a;"^^/^  —  a?"^^/*  +  2  2/. 

21.  a'^ft'^  +  4  a-^  +  3  a~h-^  by  a~^  -  4  a'^^-^  -  3  &-«. 

22.  x^-4.xi-^  +  Qx-^  hj  2x-i-^x-i-Sx-'^. 

23.  a"^  -  2  a-^n-i  +  3  a'^Ti-^  by  2  a-^ri-^  +  4  or^-^  -  6  n-^. 

24.  2  a^  -  a^a?  -  5  a^a;^  by  4  a"^aj~^  +  2  a-^a;"*  +  10  a"i 


242.    To  prove  —  =  a"*""  /or  aZ/  values  of  m  and  n. 
ft 

By  §  240,  ^  =  a-  X  a-«  =  a^-%  by  (1),  §  241. 

The  proof  of  this  result  in  the  case  where  m  and  n  are  positive  integers, 
and  m  >  n,  is  given  in  §  70.  . 

1.  Find  the  value  of  — •  - 

We  have,  «I  =  oi+2^of 

\ 


THEORY  OF  EXPONENTS 


217 


2.   Find  the  value  of 


Va' 


<^'     a^ 


3.   Divide  18  xy-^  -  23  +  x~Hj  +  6  x'Y 


by  3  x^y~'^  -\-x'^  —  2  x~*y. 


18  a;i/-2  -  23  +  X  ^y  +  6  x-'^y^ 
18  xy-2  +  6  x^y-i  -  12 


3 x'^y-'^  +  x^  -2x  ^y 


6  x^i/-i  -2x  ^-3x  ^y 


6  x^?/-i  -  11  +     X  ^?/  +  6  x-12/2 


-6xV 

1  _ 

2  +  4x' 

% 

- 

9-3x 
9-3x" 

%  +  6  X- 

%  +  6x- 

.ly2 

It  is  important  to  arrange  the  dividend,  divisor,  and  each  remainder  in 
the  same  order  of  powers  of  some  common  letter. 


EXERCISE  98 

Divide  the  following : 
1.   ar'  by  x^.  4.    m~^  by  -y/m. 


5.   a~^  by 


'^a^ 


7.  Va'  by  ■^/a-^ 

8.  S-y/mF'  by  2  m-^. 

9.  9a~^b-'hy3a'b-i 


2.  a^  by  ai  .      a, 

3.  71  by  n"i  6.    rc^  by  a;"rV 

10.  ic"' —  2  a;"' —  8  x^  by  a?"^  — 4A 

11.  a-'-b-''  by  a-^-5-l  12.   a^-1  by  x^  +  1. 


13. 


7  +  n-^  by  n^  +  S  +  n-^ 


^14.  a-^  +  4a-^-2a-^-12a-^4-9  by  a-*  +  2a-^-3 

\/15.  8m^  +  12m^n^  +  6mM4-^^  by  2m^  +  ni 

16.  a^?/-i2_;i^;^^4y-6_^l  ^y  a.V'  +  3a^7/-^-a^2/"'- 

17.  a"*-  +  2  a-^6-2  +  9  b''  by  a-^  +  2  a'h-''  +  3  a"^&-2. 


218  ALGEBRA 

18.  4  aV^  - 17  a^x^  + 16  a~^a^  by  2  a*  -  a^  -  4  a" V. 

19.  9  m*n~3  — 10  m^n^  +  m~*n  by  3  m^?i^  —  4  mn^  +  m%. 

243.  We  will  now  show  how  to  prove  equation  (2),  §  235, 
for  any  values  of  m  and  n. 

We  will  consider  three  cases,  in  each  of  which  m  may  have 
any  value,  positive  or  negative,  integral  or  fractional. 

I.  Let  ?i  be  a  positive  integer. 

The  proof  given  in  §  93  holds  if  n  is  a  positive  integer,  what- 
ever the  value  of  m. 

II.  Let  n  =  — ,  where  p  and  q  are  positive  integers. 
Then,  by  the  definition  of  §  237, 

(ay  =  -V(^^  =  ^/aFp  (§  243,  I)  =  a"^. 

III.  Let  n  =  —  s,  where  s  is  a  positive  number. 
Then,  by  the  definition  of  §  239, 

(a-)-*  =  -i-  =  —(%  243,  I  or  II)  =  a-^\ 

Therefore,  the  result  holds  for  all  values  of  m  and  n. 

1.  Find  the  value  of  (a^)-^ 

We  have,  (a2)-5  =  a^x-s  =  ^-lo. 

2.  Find  the  value  of  (a-^)"^. 

(a-3)-^  =  a"^^-^  =  a. 

3.  Find  the  value  of  (Va)l 

EXERCISE  99 
Find  the  values  of  the  following : 
1.  (a^)-*.  2.  (x-y.  3.  (x^)i  4.  (a-'yi 


THEORY  OF  EXPONENTS  219 


5.  (m-fy.         8.  (a^)-^o.  H.  (%-i)-^        14^ 


7.  (a-»)-i        "■  Vn^J'  13.  (x'y^.  16.  (a»^-V-=i; 

244.  The  value  of  a  numerical  expression  affected  with  a 
fractional  exponent  may  be  found  by  first,  if  possible,  extract- 
ing the  root  indicated  by  the  denominator,  and  then  raising  the 
result  to  the  power  indicated  by  the  numerator. 

Ex.     Find  the  value  of  (-  8)i 

By  §  243,     (-  8)^  =  [(-  8)^]2  =  (^/38)2  =  (_  2)2  =  4. 

EXERCISE  100 
Find  the  values  of  the  following : 

1.  27l        5.   81-1  9.   256-4.  13.  243-1 

2.  lel        6.    (-32)1       10.   (-512)-^.  14.  (-128)7. 

3.  64l        7.   36-1  11.   9l      ^^  15.  729-^ 

4.  64l        8.   (-216)1      12.    (-8)"^.  16.  5121 

245.  We  will  now  show  how  to  prove  the  result 

(aby  =  a^'b'', 

for  any  fractional  or  negative  value  of  n. 

The  proof  of  this  result  in  the  case  where  n  is  any  positive 
integer,  was  given  in  §  94. 

I.   Let  n  =  ^,  where  p  and  q  are  any  positive  integers. 
By  §  243,  [(abyy  =  (aby  =  a^b^  (§  94).  (1) 

By  §  94,  (Jbly  =  {a^yib^y  =  a^hK  (2) 

From  (1)  and  (2),  \_{aby  ]'  =  (a«  6 * )*. 


220  ALGEBRA 

Taking  the  qth.  root  of  both  members,  we  have 
p        p  p 
(aby  =a^h^. 

II.    Let  n  =  —  s,  where  s  is  any  positive  integer  or  positive 
fraction. 

Then,  (a&)-  =  ^  =  4^(§§  94,  or  245,  I)  =  a-6- 

EXERCISE  101 
Find  the  values  of  the  following : 

1.  {ah-y.  3.    (ic'V)^-  5.    (n-V^)-". 

2.  (m-^n-^)-\  4.    {a^x^y^-  6.    (a/o^^^^I 

^IVIISCELLANEOUS  EXAMPLES 
EXERCISE  102 
Square  the  following  by  the  rule  of  §  97  : 
^  1.   3a^  +  4  6"l  2.   5  m-^w^  -  8  m^n-*. 

3.  Square  ft^^,-^  _  2  a'  -  a,-i6^  by  the  rule  of  §  204. 

4.  Expand  {^  x^y~^ -\- 1  z-"^  {4.  x^y~^  - 1  z-^)  by  the  rule  of 
§98. 

Find  the  value  of  : 

^         25a-«-49m^    ^    .,.        ,      o  .  ... 

5.    ~,  by  the  rule  of  §  101. 

5  a-3  _  7  y^ii  • 

^    8  i«2  4-  27  ?r^ 

2  a;*  +  3  2/~*  i-       ?' )  -i      -V  /   '^ 

-  ^'  -^ z^-  ■  8-   —, 1,  by  the  rule  of  §  103. 

x^-x    3  a* +  6"^ 

9.    {3x^-4.y-fy.  10.   (a-263  +  2a«6-y. 


THEORY  OF  EXPONENTS  221 

Find  the  square  roots  of  the  following : 

11.   16 a-^mK  12.   49a;%2~l  13.    ^?     • 

4  b^n-^ 

14.  9x^-6x'i  +  25-Sx-^  +  16x-\ 

15.  4a-^4-20a-5  +  21a-^-10a-^  +  l. 

16.  a^6-3  -  6  a'6-2  +  5  ^-^  + 12  a,"*  +  4  a~^6.    • 

Find  the  cube  roots  of  the  following : 

_17.   8a^r'-        18.    -64a-'^6M.        19.   ?I^?L!^. 

x^y  2 

20.  27aj^4-54a;^2/"*  +  36a;^2/"^  +  8  2/-^. 

21.  x^ -6  x'^  +  21  x-^ -Ux-^  +  63  x-^  -54:x-^  +  27  x-^. 

Simplify  the  following,  expressing  all  the  results  with  posi- 
tive exponents : 

^.   [^(.VO^^(»-V)]«        23.     -^x^. 

■y/lf-</c  yjo? 

k24.   [a"-i  X  (a-0"+^]  x  [(a-*^)"'  x  (a"-^)-!]. 

25.   (x'^  xx^-^y.  31,   JLt^ aMi&!. 

a^  +  b^       a-6 

(^m+nX  2m  //i  2m\  m— n 
^   l^j     ■  32.    ai±6*  +  ?l±61*. 


n+1 


a^-fti     a"^-6"^ 


27.    (a'*-^  ^  a"+^)  2"  .  §!? 


1         1 


33  ^:^-i^»^^ 


mn 


28.  (a;'"'-f-ic"')'^+«-f-a;  ».  x^^'  +  l     a^"  — 1 

29.  [^(0.^-0?-    .  34.    -^--^x-j—^  +  l. 


30    ^^  +  y^      a?  4-  y .  35    g^  +  2  6^        7  a^6^  +  6b^ 


222  ALGEBRA 

0.  wa     ^       XVIII.    SURDS   ..       ^  . ;  _  f> 

246.  A  Surd  is  the  indicated  root  of  a  number,  or  expression, 
which  is  not  a  perfect  power  of  the  degree  denoted  by  the  index 
of  the  radical  sign ;  as  V2,  VS,  or  -^x  +  y. 

fK^Jix^*--^-  247.  A  monomial  is  said  to  be  rational  when  it  is  rational 
-^^--Jt^U  ^^^  integral  (§  63),  or  else  a  fraction  whose  terms  are  rational 
,  Jt3.   and  integral. 

wvT  ic  t>v^.A  polynomial  is  said  to  be  rational  when  each  of  its  terms  is 
rational. 

An  expression  is  said  to  be  irrational  when  it  involves  surds ; 
as  2  +  V3,  or  Va  4- 1  —  Va. 

248.  A  rational  number  is  a  positive  or  negative  integer,  or 
a  positive  or  negative  fraction. 

An  irrational  number  is  a  numerical  expression  involving 
surds ;  as  V 3,  or  2  +  V5. 

249.  If  a  surd  is  in  the  form  bVa,  b  is  called  the  coefficient 
of  the  surd,  and  ri  the  index. 

250.  The  degree  of  a  surd  is  denoted  by  its  index ;  thus,  V5 
is  a  surd  of  the  third  degree. 

A  quadratic  surd  is  a  surd  of  the  second  degree. 

REDUCTION  OF  A  SURD  TO  ITS  SIMPLEST  FORM 

/  251.  A  surd  is  said  to  be  in  its  simplest  form  when  the 
expression  under  the  radical  sign  is  rational  and  integral 
(§  63),  is  not  a  perfect  power  of  the  degree  denoted  by  any 
factor  of  the  index  of  the  surd,  and  has  no  factor  which  is  a 
perfect  power  of  the  same  degree  as  the  surd. 

252.  Case  I.  When  the  expression  under  the  radical  sign  is 
a  perfect  power  of  the  degree  denoted  by  a  factor  of  the  index. 

Ex.    Reduce  V8  to  its  simplest  form. 

We  have,  \/8  =  v^  =  2  "^  (§  2.37)  =  2^  =  \/2. 


SURDS  223 

'Exercise  103 

Eeduce  the  following  to  their  simplest  forms : 

1.  </2E.  5.    a/49.  9.    ^"243.  13.  a/216  aV. 

2.  <m.  6.    ^^.        10.    '</U3.  14.  a/64^^. 

3.  v/i2i.        7.    a/64.        11.    ■v^l44^.        15.  ^^8^^^ 


9/T777^  o        10 


125.        8.    V81.        12.    V27nV.         16.    V625^. 

253.  Case  II.  TF/ie)2  the  expression  under  the  radical  sign  is 
rational  and  integral,  and  has  a  factor  which  is  a  perfect  power 
of  the  same  degree  as  the  surd. 

1.   Reduce  V54  to  its  simplest  form. 

We  have,      \/54  =  v/27"x2  =  v^  x  ^  (§  234)  =  3^2, 


2.   Reduce  V3  a^b  - 12  a^b^  + 12  ab^  to  its  simplest  form. 


V3  a^b  -  12  a262  +  12  aft^  =  y  (a2  _  4  a&  +  4  b'')S  ah 
=  Va2  -  4  a6  +  4  62>/3a&  =  (a  -  2  6)  VSoft. 

We  then  have  the  following  rule : 

Resolve  the  expression  under  the  radical  sign  into  two  factors, 
the  second  of  which  contains  no  factor  which  is  a  perfect  power  of 
the  same  degree  as  the  surd. 

Extract  the  required  root  of  the  first  factor,  and  multiply  the 
result  by  the  indicated  root  of  the  second. 

If  the  expression  under  the  radical  sign  has  a  numerical 
factor  which  cannot  be  readily  factored  by  inspection,  it  is 
convenient  to  resolve  it  into  its  prime  factors. 

3.   Reduce  Vl944  to  its  simplest  form. 


4.   Reduce  Vl25  x  147  to  its  simplest  form. 
Vl25  X  147  =  V68,x  3  X  72  =  V52  x  72  x  y/bxZ  =  5  x  7  X  Vl5  =;  35yi6. 


224 


ALGEBRA 


^  EXERCISE  104 
Reduce  the  following  to  their  simplest  forms : 

1.  V90.         5.   ^56.              a^l9^.  13.  V242ar'/. 

2.  V72.      ""6.  7VW.  ^'^10.  a/432.  ^  14.  -^SOOa^ftV. 

3.  V96.         7.  9^/81.  '      11.   ^256.  15.  a/162  m^n^^ 

4.  a/75.         8.  a/48.           12.  V500  a^6^  16.   ^160  ^fz\ 
17. 


18. 


23. 
24. 
25. 


V75ar't/8-100a;Y. 
a/128  a^63+320  a%\ 


19.  A/(3a;  +  2?/)(9ar2-4  2/'). 

20.  A/3a3-24a2+48a. 


N 


21.  a/18  a^6  +  60  a^i^  +  50  a6^. 

22.  a/(2  «2  +  a;  - 15)(2  a^  -  19  a;  +  35). 


V896. 
a/2268. 


26.    a/98  X  196. 


27.    V432x504. 


V5145. 


28.    V1372. 


29.  a/7875. 

30.  a/375  X  405. 
^31.    a/54  X  63x336. 


32.    V63  xf  X  175  2/2'  x  875  ^o^. 


254.  Case  III.  Wlien  the  expression  under  the  radical  sign 
is  a  fraction. 

In  this  case,  we  multiply  both  terms  of  the  fraction  by  such  an 
expression  as  will  make  the  denominator  a  perfect  power  of  the 
same  degree  as  the  surd,  and  then  proceed  as  in  §  253. 

Ex.     Reduce  \/- — -  to  its  simplest  form. 
^8a^ 

Multiplying  both  terms  of  the  fraction  by  2  a,  we  have 


EXERCISE  105 
Reduce  the  following  to  their  simplest  forms : 
1.   A/f         2.    Vf.  3.    a/^-  4.    a/||. 


5.   Vf|. 


SURDS  225 


^7.  ■^.;s'^^"12.  -v^Y- 


/  23 
*    ^32a^* 


25  6 


3/T  13.   V|.  ly    ^/llaV  20.  ^^^. 

q      3/13  '       -H  , , 

^  '15    J^        \18    A^^  21    ^C^^. 

^22  a;^  /3  a^  - 18  eg  +  27 

255.  To  Introduce  the  Coefficient  of  a  Surd  under  the  Radical 
Sign. 

The  coefficient  of  a  surd  may  be  introduced  under  the 
radical  sign  by  raising  it  to  the  power  denoted  by  the  index. 

Ex.  Introduce  the  coefficient  of  2V3  under  the  radical 
^^^^'  2^3  =  v^  X  ^3  =  VISITS  (§ 234)  =  ^M. 

A  rational  expression  (§  247)  may  be  expressed  in  the  form  of  a  surd 
of  any  degree  by  raising  it  to  the  power  denoted  by  the  index,  and  writ- 
ing the  result  under  the  corresponding  radical  sign. 


i^  EXERCISE  106 

Introduce  the  coefficients  d^B^following  under  the  radical 
signs :  ^jp,, 

1.  3V7.  3.   4a/5.  5.   4v'5.  7.   2^/3. 

2.  QVQ.  4.   5^/7.  6.   2v'8.  8.   9xV2^. 
9.    lOa^ftVe^. 


!*•  (2»+i)Vj=E3- 


10.  6»/-v'T?. 

11.  5a»«\/2A 

12.  Za?V^/Zl^.  '"    «  +  l^a*-4a  +  3 


11.   5  an?-i/2  a'n.  „     g-1    /g^  +  3  a  +  2 

"•    a  +  lVa^_4a 


13.    /„      .^./S+^.  ifi     g-^-/o      (=^-2)' 


'•  («-^)\&|-  !«•  ^W^ 


(^-1)' 


226  ALGEBRA 

ADDITION  AND  SUBTRACTION  OF  SURDS 

256.  Similar  Surds  are  surds  which  do  not  differ  at  all,  or 
differ  only  in  their  coefficients ;  as  2 V<xa.*^  and  S^vax^- 

Dissimilar  Surds  are  surds  which  are  not  similar. 

257.  To  add  or  subtract  similar  surds  (§  256),  add  or  sub- 
tract their  coefficients,  and  multiply  the  result  by  their  common 
surd  part. 

1.  Required  the  sum  of  V20  and  ViS.  * 
Reducing  each  surd  to  its  simplest  form  (§  253), 

V^  +  V45  =  V4^^  +  V9ir5  =  2  VS  +  3  V5  =  5  Vs. 

2.  Simplify  VJ  +  V|-V|. 

=:iV2+lV6_§V^  =  lV6-lV2. 
2  3  4  3  4 

We  then  have  the  following  rule : 
Reduce  each  surd  to  its  simplest  form. 

Add  or  subtract  the  similar  surds,  and  indicate  the  addition  or 
subtraction  of  the  dissimilar. 

EXERCISE  107 

Simplify  the  following : 
^l.   V8+V32.  3.   V300-V147.  5.   a/T35-^40. 

2.   V28-V63.  4.   a/2 -j- ^128.  6.   ^80  +  ^405. 

7.   V3  4-V192-V243.  8.    V250  -  V90  -  VlTG. 

9.  V|  +  V«^      ^  V^  +  Vf.  wll.  ^1-^^. 

12.    V99-V275  4-V396.  14.   V^-VV  +  VV-. 

Ms.   a/56 4- -v/189-f a/162.  15.    V||+V||-V^. 

16.   V72^-a;V98«3  4.a;V200«. 


SURDS  227 


17.   a V80  a'b'  +  ab  V270  a'b'  +  b' V640  a'b. 


1-18.    V27^M-36»V  +  V48ic/  +  64^3. 
21.    ■v/l28  +  ^250--\/432-^88. 


^  22.  V50  a'  +  V72  6^  _  V50  a^  + 120  a6  +  72  d^. 

^23.  ^96  + ^486 -a/6. 

24.  V294-V216+V405-V600. 

^25.  V52^2_5yjX7^_y'j26a^  +  aV56a^. 

26.  VI  +  VI-VS-VS.      27.  VII-VS-VI+V?. 


28.    V50a3«  +  40a;2_f_8^_y32aj3_48x2  +  18x. 


y  29.   V125 x^ - 150 a;?/ +  45 /  + V5a;2  + 60 icy  + ISO/. 

^x  —  y  ^x-\-y        x^  —  y^ 

TO  REDUCE  SURDS  OF  DIFFERENT  DEGREES  TO  EQUIVA- 
LENT SURDS  OF  THE  SAME  DEGREE 

258.  Ex.  Reduce  V2,  ^3,  and  ^5  to  equivalent  surds  of 
the  same  degree. 

By  §  237,  V2  =  2^  =  2X2  =  '^26  =  ^/64. 

We  then  have  the  following  rule : 

Express  the  surds  with  fractional  exponents,  reduce  these  to 
their  lowest  common  denominator,  and  express  the  resulting 
expressions  with  radical  sigyis. 

^  /  The  relative  magnitudes  of  surds  may  be  determined  by  reducing  them, 
\yE^necessary,  to  equivalent  surds  of  the  same  degree. 

Thus,  in  the  above  example,  \/l25  is  greater  than  \/8l,  and  Vsl  than 

\    Then,  y/b  is  greater  than  \/3,  and  Vz  than  V2. 


228  ALGEBRA 


EXERCISE  108 


Reduce  the  following  to  equivalent  surds  of  the  same  degree : 


ca. 


^1.  V2  and  V7.  6.    -</ah,  -Vbc,  and  V 

v2.  V3  and  a/4.  ^.   V2^,  ^3b,  and  -v^Sc. 

3.  a/2  and  ■^/3.  8.   a/2^,  "v/S/,  and  -y/W?. 

\  4.  -v^S  and  a/5.  9.   a/^ThI  and  ^^^^. 

5.  a/3  and  \/6.  10.   ->Jx  —  y  and  \/aJ  +  2/. 

11.  Which  is  the  greater,  VG  or  a/14? 

^  12.  Which  is  the  greater,  a/2  or  -^5  ? 

13.  Which  is  the  greater,  a/3  or  a/7  ? 

14.  Arrange  in  order  of  magnitude  a/2,  a/13,  and  a/31. 

15.  Arrange  in  order  of  magnitude  a/4,  a/6,  and  a/15. 
Arrange  in  order  of  magnitude  a/2,  a/3,  and  a/10, 

MULTIPLICATION  OF  SURDS 
259.   1.  Multiply  V6  by  VlS. 
By  §  234,  \/6  X  \/l5  =  VCxlS  =  \/2x3x3x6  =  V32x2x5  =1  VlO.    . 

2.   Multiply  a/2^  by  -y/^K  ^^0 

Reducing  to  equivalent  surds  of  the  same  degree  (§  258), 

V2a  X  y/Tcfi  =  (2  a)^  x  (4  a^)^  =  (2  ay  x  (4  a^y  =  v^(2a)8  x  v^(4  a'^Y 

=  v^28  a8  X  24  a*  =  v/26  a6  X  2  a  =  2a\/Ui. 

We  then  have  the  following  rule: 

Tb  multiply  together  two  or  more  surds,  reduce  them,  if  neces- 
sary, to  syrds  of  the  same  degree. 

^ultij^Jtog ether  the  expressions  under  the  radical  signs,  and 
write  the  result  under  the  common  radical  sign. 

The  result  should  be  reduced  to  its  simplest  form. 


SURDS 


229 


3.  Multiply  V5  by  -y/B. 
By  §  237,  V5  =  6^  =  6^  =  y/bK 

Then,  V5xv'5  =  v'Px^  =  \^  =  5^  =  5^  =  \^=v^. 

4.  Multiply2V3  +  3V2by3V3-V2. 

2V3  +  3V2 
SVS-    V2 

18  +  9\/6 
-2\/6-6 

18  +  7V6-6  =  12  +  7V6. 

To  multiply  a  surd  of  the  second  degree  by  itself  simply  removes  the 
radical  sign  ;  thus,  \/3  x  V3  =  3. 

5.  Multiply  3 VlT^  -  Wx  by  VI  +  a;  +  2V^. 


3vr+x-4Vx 
vT+^+2Vi 


3(l  +  x)-4VxT^ 

+  6\/x  +  x2-8x 


8(1  +  x)  +  2  Vx  +  x2  _8x  =  3-6a;  +  2  Vx  +  a;^. 


EXERCISE  109 


Multiply  the  following : 
j   1.    V5  by  V20. 
\?.    ^/49¥2  by  VTx. 

3.  Vl5  by  V27. 

4.  Vi8  by  V42. 


10. 

11. 

12. 

-13. 

v^5.    VlOSby  Vl92.ll^*^^^14. 
\s.   ^72  by -^81.  ,>^  3:^15. 

16. 
17. 
18. 


V?by  V|f_ 
^/98  by  a/343. 
-\/63by 


7.  -\/55xy  by  V66  2/2. 

8.  VsEhy-y/TE. 

9.  a/84  by  a/ISO. 


''-  '     -^135. 


V6by  A^. 

V3  »2/  by  -y/Tyz. 
-s/M  by  A/i2. 
a/135  by  ^45. 
V20  by  ^. 


230  ALGEBRA 


V 19.    V5d'  by  -^125^.  ^  23.   -s/^,  -Vyi,  and  ^^. 

20.  -v/Q  by  -^27.  24.    V20,>25,  and  ^5. 

21.  VS  by  ^^,.  25.    -Vl,  ^6,  and  -J/6. 

^  22.    V|  by  ^S-  ^26.    Vi5,  -^^,  and  ^^. 

x>27.   6 +  3V2  and  4 +  5V2. 

28.   4Va-3V6and7Va  +  2V6. 
^29.   2V5-8^/3and9V5-4V3. 

30.  5V2  +  6V6andl0V2-7V6. 

31.  2a/9  +  9^7  and  8-^-3^49. 

32.  4V^-V^  +  3\/^and4VicH-Vy-3V^. 

33.  3 Va  +  2 H- 4 Vci - 1  and  6 Va  +  2  +  ^^a^^, 
^34.    V2+V5+V7and  V2-V5-V7. 

/35.   4V|-3VSand2V|-9V^. 

36.  3V3  +  2V6-4V8and3V3-2V6  +  4V8. 

37.  6V5  -  5V7-VI0  and  6V5  +  5V7  4- ViO. 
1^38.   8Vl2  +  7V20-4V24and5V3-3N/5  +  2V6. 

,39.   6V|  +  8V|  +  llV|and3V|-4V|-5V|. 


DIVISION  OF  MONOMIAL  SURDS 
260.   By  §234,  ■\/^  =  VaxV5. 


Whence,  2U^^-^, 

'  n/— 

We  then  have  the  following  rule : 

To  divide  tvjo  monomial  surds,  reduce  them,  if  necessary,  to 
surds  of  the  same  degree. 

Divide  the  expression  under  the  radical  sign  in  the  dividend  by 
the  expression  under  the  radical  sign  in  the  divisor,  and  write  the 
result  under  the  common  radical  sign. 

The  result  should  be  reduced  to  its  simplest  form. 


SURDS 


231 


1.   Divide  7405  by  </5. 


We  have, 


v^iOS 


»    Ft 


^5        ^  5 

2.   Divide  v^  by  V6. 

Reducing  to  surds  of  the  same  degree  (§  258), 

4^ 


^4 


mi 


VQ     qI      (2x3)^      </W^^^ 

3.  Divide  VlO  by  -^40. 


/     2^ 
'23x33 


f=^=i^. 


"We  have. 
Then, 


VIo  =  10^  =  10^  =  v^Io3  =  V(2>rEy. 


^40      ^  23  X  6 


\ 


Divide  the  following : 

1.  V90  by  V5.         3. 

2.  V24  by  Vl8.       4. 

7.  </32  by  -\/2. 

8.  a/186  by  •^. 
V9.    V7  by  v'lQ. 

10.    ^42^  by  ^56^. 
a/686  by  ^63. 
V25^  by  -^25^. 
VS  by  Vf|. 


EXERCISE  110 

VTO  by  V  63 
^144  by  </9. 


11. 
12. 
13. 
14. 

15.   </M  by  ^S- 


l^V  by  V5f 


5.  </3  by  a/192. 

6.  a/48  by  -^56. 


16.    V6ab'  by  ^96  ftc^, 
\/3a^  by  V2a. 


\7 

18. 

19. 
v^20. 

21. 

22. 

23. 


V27^  by  -^36^. 
V|  by  7||. 

<m  by  V|. 

\/12V5^  by  a/4^^. 

by  ^. 

_.  by  ^. 
24.    a/15^2  by  -^/ioS^. 


INVOLUTION  OF  SURDS 
261.   1.   Kaise  ■V^12  to  the  third  power. 

(%^)«  =(12^)8  =  12^  (§  243)  =  12^  =\/i2  =  2V3. 


232  ALGEBRA 

2.  Raise  V2  to  the  fourth  power. 

(  ^2)*  =  (2^)4  =  2^  =  v^  =  v^. 
Then,  to  raise  a  surd  to  any  positive  integral  power, 
I       If  possible,  divide  the  index  of  the  surd  by  the  exponent  of  the 
required  power  ;  otherwise,  raise  the  expression  under  the  radical 
sign  to  the  required  power. 

The  rules  of  §  §  97  and  98  should  be  used  to  find  the  value  of 
any  product  which  comes  under  them. 

3.  Expand  ( V6  -  VS)^. 

By  §97,         (V6-\/3)2  =  (\/6)2-2\/6x  V3+(V3)2 
=  6  -  2\/PV2  +  3  =  9-  Q^/2. 

4.  Expand  (4  +  -^)  (4-^). 

By  §  98,  (4+  v^)(4-  Vb)  =42-  (^)2=16-  ^,  by  the  above  rule. 

%^_^\      EXERCISE  III 

Find  the  values  of  the  following :  * 

1.    (^/2)*.                    6.   (5^5«^)2.  vll.  (^a^^/2E^y,^^ 

/2.   (-5^6)2.                    7.    (•v/^'^)^  12.  (4^^729)3. 

3.  (^4a;-f  3  2/)l     ^8.   (^72^^)^  13.  (7  +  2V2)2. 

4.  (^32)1     Z :          9.    {VU^f.  V14.  (4V5-5)2. 
.b.    (V2^^y.            10.   {-y/ly.  Vl5.  (3V6  +  6V3)^ 

16.  (9V7-4VIl)l  -^18.   (4V^^:^  +  3V^T6)'. 

17.  (V5a;  +  2-V3^)2.  Vl9.    (6  4-5V2)(6-5V2). 
v^20.   (4V^  +  3V^^^)(4Va-3V^:^^). 

'/  21.  (V2a;  +  2/  +  V2a;-2/)(V2a;  +  2/-V2a;-y). 

22.  (5V3ic4-4:  +  4V5a;-2)(5V3a;  +  4-4V5a;-2). 

^23.  ^4  +  2V3x^4-2V3.  ^  24.    (^  +  ^9)(v^4-\/9). 

>  25.  V3V5  +  2V7  X  V3V5-2V7. 

26.  Expand  (2  V2  +  V6  -  V3)2,  by  the  rule  of  §  204,, 


SURDS  233 

EVOLUTION  OF  SURDS 

262.  1.   Extract  the  cube  root  of  \/27a^. 

\/(  v/27x8)  =  (  \/(3^P  =  [(3  x)^]*  =  (3  cc)^  =  \/3«. 

2.   Extract  the  fifth  root  of  V6. 

V{  \/6)  =  (6*)^  =  6^^  =  v^. 

Then,  to  extract  any  root  of  a  surd, 

\  If  possible,  extract  the  ^^equired  root  of  the  expression  under  the 
radical  sign  ;  otherwise,  multiply  the  index  of  the  surd  by  the  index 
of  the  required  root. 

If  the  surd  has  a  coefl&cient  which  is  not  a  perfect  power  of  the  degree 
denoted  by  the  index  of  the  required  root,  it  should  be  introduced  under 
the  radical  sign  (§  255)  before  applying  the  rule. 

Thus,       '  v/(4V2)  =  ^(V32)  =  \/2. 

\^,_,^S_  EXERCISE  112 

Find  the  values  of  the  following :  ^  a,  'iT  -. 

1.  V(>^y       ^/5.  V(^/9a2+12a  +  4).      9.  -Vil^a'-s/^). 

2.  V(^i3).         v6.  a/(a/49).    ^7  ^10.  V(2xV^y 

3.  ^{VW^y     i  7.  a/(81-^I6).  3  a  11.  </(^3^). 

v4.  ^/(■</24S^),      8.  ^/(2^/3a^).  12.  's/(2n'Vie^'). 

\ 

REDUCTION    OF    A    FRACTION    WHOSE    DENOMINATOR    IS 

IRRATIONAL   (§  247)   TO  AN  EQUIVALENT  FRACTION 

HAVING  A  RATIONAL  DENOMINATOR 

263.  Case  I.     When  the  denominator  is  a  monomial. 

The  reduction  may  be  effected  by  multiplying  both  terms  of 
the  fraction  by  a  surd  of  the  same  degB*e  as  the  denominator, 
having  under  its  radical  sign  such  an  expression  as  will  make     \ 
the  denominator  of  the  resulting  fraction  rational. 


234  ALGEBRA 

Ex.    Keduce  — ==  to   an   equivalent  fraction    having   a 

rational  denominator. 

Multiplying  both  terms  by  Vq  a,  we  have 

5      ^     6</9a      _  5</9a  ^  5v^9g. 


EXERCISE  113 

Keduce  each,  of  the  following  to  an  equivalent  fraction  hav- 
ing a  rational  denominator ; 

1.   A.  3.   -^.  5.      ^  -       ^ 


VE'  </6a'  V25  V27 

2  2      .        4    -4=.  6.   -4^.       8.         ^      . 

264.   Case  II.      Whe^i  the  denominator  is  a  binomial  contain- 
ing only  surds  of  the  second  degree. 

1.  Eeduce      ~     _   to    an    equivalent   fraction    having    a 

rational  denominator. 

Multiplying  both  terms  by  5  —  V2,  we  have 

5_V2^         (5-V2)2        ^25-10V2  +  2  ..gg^  ggN^^T-lOV^ 
5  +  V2      (6+V2)(6-V2)  25-2  '     ^  23        ' 

2.  Reduce        ^^    Vq—     ^^  ^^^  equivalent  fraction  having 

2Va  — 3Va— 6 
a  rational  denominator. 

Multiplying  both  terms  by  2\/a  +  3  Va  —  b, 

3Va  -  2Va  -  6  ^  (3v^  -  2 V^"^r6)(2v^  +  SV^i^^) 
2y/a  -  SVa  -  b     (2Va  -  3V^"I^)(2Va  +  3Va^^) 


6  q  +  SVa  Vo"^^  -  6(a  -  6)  _  6  &  +  SVa^  -  a6 
4a-9(a-6)  9b -5a 


SURDS  235 

We  then  have  the  following  rule : 

Multiply  both  terms  of  the  fraction  by  the  denominator  with  the 
sign  between  its  terms  reversed. 

EXERCISE  114 

Keduce  each  of  the  following  to  an  equivalent  fraction  hav- 
ing a  rational  denominator : 

^         8                     ^  "V^+ Vy.  i^  7^    3V5-V3 

V6  +  2                   *    V^-V^  *   4V5  +  5V3 

2           7    -                 5     V10-6V2  ^ g    3-Va^ 

5-3V2  ■             *    VIO  +  2V2  *   4-Va^ir3 


Vti.  ^g^    2V7  +  3V3,        ,Q     Va;-Va;  +  y 


m  +  Vn                      2V7  — 3V3                 VaJ  +  Vaj  +  y 
jQ     V9a^-2-3a  13.  ?_ 


V9a^-2  +  3a 


11.    Va^  +  /  +  y. 


14. 


VVll  +  3-VVll- 

-3 

Va^  +  2/2  +  Var^- 

-f 

Vic2  +  2/2-Va:2- 

-f 

Va-fl4-2Va- 

-1 

2Va;-2-Va;  +  2  *   4 Vo  +  l  +  3 Va^=3 

265.  If  the  denominator  is  a  trinomial,  containing  only  surds 
of  the  second  degree,  the  fraction  may  be  reduced  to  an  equiva- 
lent fraction  having  a  rational  denominator  by  two  applications 
of  the  rule  of  §  264. 

Ex.     Reduce  —^ — _~         to  an  equivalent  fraction  having 

4  +  V3-V7 

a  rational  denominator. 

Multiplying  both  terms  by  4  +  \/3  +  \/7,  we  have 

4-V3.^V7^(4-\/8-\/7)(4  +  V3  +  V7)^  42-(V3  +  V7)2   .^qq. 
4  +  V3-\/7      (4  +  V3-V7)(4  +  \/3  4-V7)     (4  +  V3)2 -(V7)2 


236  ALGEBRA 

Then    4  -  VS  -  \/7  ^  16  - (10  +  2 \/2l)  ^  6  -  2\/21  ^  3  -  V21 
4  +  V3-V7         19  +  8\/3-7         12  +  8\/3     6  +  4V3 

Multiplying  both  terms  of  the  latter  by  6  —  4\/3, 
4  _  V3  _  V7  ^  (3  -  V2T)  (6  -  4  V3) 
44.V3-V7  62-(4V3)2 

_  18  -  6  V2I  -  12\/3  +  4  V63  ^  -  9  +  3V2l  +  6\/3  -  6V7 
-  12  6  ' 

The  example  may  also  be  solved  by  multiplying  both  terms  of  the  given 
fraction  by  4  -  \/3  +  V7,  or  by  4  -  >/3  -  VT^. 

EXERCISE  115 

Eeduce  each  of  the  following  to  an  equivalent  fraction  having 
a  rational  denominator : 

1.   I 3.  12 


2+V2  +  V3  V5_V3-V2 

2  6  4     V6+V3-3V2 

3+V5-V2*  .    *   .V6-V§  +  3V2' 

The  reduction  of  a  fraction  having  an  irrational  denominator  to  an 
equivalent  fraction  having  a  rational  denominator,  when  the  denominator 
is  the  sum  of  a  rational  expression  and  a  surd  of  the  nth  degree,  or  of  two 
surds  of  the  nth  degree,  will  be  found  in  §  446. 

266.  The  approximate  value  of  a  fraction  whose  denominator 
is  irrational  may  be  conveniently  found  by  reducing  it  to  an 
equivalent  fraction  with  a  rational  denominator. 

Ex.    Find  the  approximate  value  of to  three  places 

of  decimals.  2  —  V2 

1       ^  2+'-\/2  ^  2+V2  _  2  +  1.414...  ^  J  ^^^  ^^^ 

2_V2      (2-V2)(2+\/2)       4-2  2 

EXERCISE  116 

Find  the  values  of  the  following  to  three  places  of  decimals : 

^- 1.  A.  .  2.      ^  «       1 


V6  3-V3  5  +  2V7 


a    , ,  SURDS  237 

4.    _A_.  ^6.^  V7-V2  g    4V5-5V3 

■-^•v/49  '    V7  +  V2  '    4V5  +  5V3* 

/5           23  ,7     2V6  +  V3  g    4  V7  +  7  V3 

_<^    '    V5-3V2'  *    2V6-V3'  '    3V7-5V3* 

/^        PROPERTIES  OF  QUAliliATIC  SURDS  (§250) 

267.   A  quadratic  surd  cannot  equal  the  sum  of  a  rational 
expression  and  a  quadratic  s^ird. 

For,  if  possible,  let         Va  =  6  +  Vc, 

where  6  is  a  rational  expression,  and  Va  and  Vc  quadratic 
surds. 

Squaring  both  members,   a  —  ¥  +  2h^c-\-c, 

or,  2  &  Vc  =ia  —  h'^  —  c. 


Whence,  Vc  = 


g  —  0^  —  c 
26 


That  is,  a  quadratic  surd  "equal  to  a  rational  expression. 
But  this  is  impossible ;  whence,  Va  cannot  equal  h  +  Vc. 

268.    i/"  a  +  V6  =  c  +  V<^,  where  a  and   c  are  rational  ex- 
pressions, and  V&  and  's/d  quadratic  surds,  then 

a  =  c,  and  V&  =  Vc?. 

If  a  does  not  equal  c,  let  a=:c-}-x;  then,  x  is  rational. 
Substituting  this  value  in  the  given  equation, 

c-\-x  +  V6  =  c  +  Vd,  or  ic  +  Vb  =  V^. 

But  this  is  impossible  by  §  267. 
Then,  a  =  c,  and  therefore  V&  =  V^. 


269.   If  V  a  +  V6  =  V^  +  Vy,  where  a,  b,  x,  and  y  are  rational 
expressions,  then  Va  —  V^  =  Va;  —  V^. 

Squaring  both  members  of  the  given  equation, 
a  +  V6  =  a;  +  2  VaJ2/  +  y. 


238  ALGEBRA 

Whence,  by  §  268,         a:=x-\'yf 
and  V6  =  2^/xy. 

Subtracting,  a  —  V6  =  x  —  2^'xy  +  y. 

Extracting  the  square  root  of  both  members, 

V  a  _  V6  =  Vaj  —  V^. 

270.   Square  Root  of  a  Binomial  Surd. 

The  preceding  principles  may  be  used  to  find  the  square 
roots  of  certain  expressions  which  are  in  the  form  of  the  sum 
of  a  rational  expression  and  a  quadratic  surd. 

Ex.     Eind  the  square  root  of  13  —  V160. 


Assume,  ^13  -  VIM  =  y/x-Vy.  (1) 

Then,  by  §  269,  ^\^+V)M  =  Vx  +  Vy.  (2) 


Multiply  (1)  by  (2),  V169  -  160  =  x~y.                                    (§  98) 

Or,  x-y  =  Z.                 •                               (3) 

Squaring  (1),  13  -  \/l60  =  a:  -  2  Vxy  +  y. 

Whence,  by  §  268,  x  +  y  =  12>.                                             (4) 

Adding  (3)  and  (4),  2  x  =  16,  or  x  =  8. 

Subtracting  (3)  from  (4),           2  y  =  10,  or  y  =  6. 

Substitute  in  (1) ,  ^IZ-y/lim  =V% -Vb  =  2^/2 -y/l. 

271.  Examples  like  that  of  §  270  may  be  solved  by  inspec- 
tion, by  putting  the  given  expression  into  the  form  of  a  tri- 
nomial perfect  square  (§  111),  as  follows : 

Reduce  the  surd  term  so  that  its  coefficient  may  be  2. 

Separate  the  rational  term  into  two  parts  whose  product  shall  he 
the  expression  under  the  radical  sign  of  the  surd  term. 

Extract  the  square  root  of  each  part,  and  connect  the  results  by 
the  sign  of  the  surd  term  (§  112). 

1.   Extract  the  square  root  of  8  +  V48. 
We  have,  •  \/48  =  2\/l2. 


SURDS  239 


We  then  separate  8  into  two  parts  whose  product  is  12. 
The  parts  are  6  and  2  ;  whence, 


V8  +  V48  =  V6  +  2  Vl2  +  2  =  V6  +  V2. 

2.    Extract  the  square  root  of  22  -  3  V32. 

We  have,  3\/32  =  V9  x  8  x  4  =  2V72. 

We  then  separate  22  into  two  parts  whose  product  is  72. 
The  parts  are  18  and  4  ;  whence, 


V22Z3V32  =  V 18  -  2  V72  +  4  =  \/l8  -  V4  =  3  V2  -  2. 

Vh  ^^ 

%0  EXERCISE  117 

Find  the  square  roots  of  the  following : 

^1.   15  +  2V54.  7.   30-V500.  13.  45-5V80. 

2.  21-2V80.  8.   13  +  VI68.  14.  34  4-12V8. 

3.  53-2V52.             9.   24  +  2ViiO.  15.  61  +  28V3. 
>4.   23  +  6VIO.  10.   44-4V72.  16.  53-V600. 
>5.   38-10V13.  11.   55-20V6.  17.  60-5VI08. 
>6.   29  +  2V54.  12.  55  +  3V24.  18.  54-h3Vi28. 


19.   4a-2V4a2-9.       20.   4.(2x-y)-\-2Vl5ay^-12xy. 

Solution  of  Equations   having  the  Unknown   Numbers  under 
Radical  Signs. 

272.   1.   Solve  the  equation  Vo^  — 5  — a;  =  — 1. 


Transposing  —  cc,  Vx^  —  5  =  x  —  1. 

Squaring  both  members,     x^  —  6  =  x^  —  2x  +  1. 

Transposing,  2x  =  6;  whence,  x  =  S. 

(Substituting  3  for  x  in  the  given  first  member,  and  taking  the  positive 
value  of  the  square  root,  the  first  member  becomes 

Vo^Ts  _3  =  2-3  =  -l; 

which  shows  that  the  solution  oj  =  3  is  correct.) 


240  ALGEBRA 

We  then  have  the  following  rule : 

Transpose  the  terms  of  the  equation  so  that  a  surd  term  may 
stand  alone  in  one  member;  then  raise  both  members  to  a  power 
of  the  same  degree  as  the  surd. 

If  surd  terms  still  remain,  repeat  the  operation. 

The  equation  should  be  simplified  as  much  as  possible  before  perform- 
ing the  involution. 

2.   Solve  the  equation  V2a;  — l-f  V2~a+~6  =  7. 


Transposing  V2a;  —  1,         V2  cc  +  6  =  7  —  \/2  x  -  1. 


Squaring,  2a;  +  6  =  49-14V'2a;-l+2x-l. 


Transposing,  14  \/2  x  —  1  =  42,  or  v  2  x  —  1  =  3. 

Squaring,  2  x  —  1  =  9 ;  whence,  cc  =  5. 

1 


3.    Solve  the  equation  VaJ  —  2  —  Va;  = 


Va?-2 


Clearing  of  fractions,     x  —  2— Vic^  —  2x  =  l. 
Transposing,  —  Va;"^  —  2  a;  =  3  —  x. 

Squaring,  x^  —  2  cc  =  9  —  6  x  +  a;2. 

9 

Transposing,  •   4  x  =  9,  and  x  =  -. 

9  4 

(If  we  put  X  =  -,  the  given  equation  becomes 

If  we  take  the  positive  value  of  each  square  root,  the  above  is  not  a  true 
equation. 

But  a  square  root  may  be  taken  as  either  positive  or  negative  ;  and  if 

we  take  the  negative  value  of  -v  - ,  and  the  positive  value  of  -v/ - ,  the  first 

1    3 
member  of  (1)  becomes -,  or  —2,  and  the  second  member  becomes 

1  ^    ^        9 

—     or  —  2  ;  then  the  3olution  x  =  -  is  correct.) 


4.    Solve  the  equation  V2  —  3a;  +  Vl4-4a;  =  V34-a;. 


SURDS  241 


Squaring  both  members, 


2-3x  +  2V2-3xVl  +  4x  +  l  +  4a;  =  3  +  x. 

Whence,  2  V2  -SxVl  +  4x  =  0 ; 

or,  *  V2-3x\/l  +4x  =  0. 

Squaring,  (2  -  3x)(l  +  4a;)  =  0. 

2 
Solving  as  in  §  126,  2-3x  =  0,  orx  =  -; 

and  H-4x  =  0,  or  x  =  -|. 

EXERCISE  118 

Solve  the  following  equations : 


^1.  V4ic  +  l+5  =  0.  V5.  V«  +  Va;  +  9  =  -2. 


2.  -\/7a;-8-2  =  -6.  6.  V37-2-V3T=l. 


3.  ■Vl6x'-\-l-4.x  =  3.  7.  V^Tl3-Va;-5  =  3. 


v4.  VSa^-{-36x'-3  =  2x.      V  8.  V5a;-19- V5a;+14=-3. 

yo  ^ ^^      =V3-2fl;. 

Va;  +  4      V3-2a; 


^    10.        ,  -=:  =  -:'      .11.    ViC— 5  +  Va;  = =:• 

■VSx-\-2-V3x     4  ^  ^^ 


Vl2.    V6a;-V6x-ll  = 


V6a;-ll 


10 


13.    V2s-V2s  +  5  =  -      

V2S4-5 

4/14.    Vx^ -Va?  +  21  =  -2 -Vx.      "^cu^^ 

3VTT2^  +  4       VH-2a^  +  6 


V  15. 


6Vl4-2a;-l     2VH-2 


X 


3V^_+4^3V^+_5^        ^^^     ■y/2a-x-h^2a  +  x_^ 
5V^-2     5V^-3  '    V2a-a;-V2aH-a; 


242  ALGEBRA 


18.    V2ri  — ic  — Vn  — fl7  =  3Vw. 
J119.    Va^-f-7a;-4  +  Va^-3a;  +  l  =  5.       V/.->v  r.-f.-  ^— 

20.    Va;-2a4-V^=        ^ 

Va;  —  2  a 


\/21.    Va;  4-  a  +  V a;  +  2  a  =  V4  a;  +  5  a. 


22.    Va-a;+V6-a;  =  V2a  +  26. 


23.    Va;  4-  V5  -  2  a;  =  V5  —  a;. 


24.  V4a;-3-V3x-l=V7a;-4. 

25.  V4i)  + 1  -  Vp  -  8  =  V9i?  -  83. 

26.  Va;  —  2  a  —  Va;  —  6  a  =  2 Va;  —  5  a. 

27.  V(3 a  +  ^/Sax-j- x^=-Vx-VSa. 

28.  V»+V4a4-aj=V464-4a?. 


29.    V2  aa;  +  6  +  V2  aa;  -  6  =  2  V2  ax  -  3  &. 


'30.    Va  —  a;  -f  V6  —  a;  =  Va  +  6  —  2  a;. 

31.  V2a;4-5a  +  V3a;  +  4&  =  V5aj4-5a  +  4  6. 

32.  V2a;-l+V3a:  +  2=V3a;-2-|-V2a;  +  3. 


33.    V2a;4-5  +  V3a5-2=V(5a;  +  3+V24a;2_^15). 

IMAGINARY  NUMBERS    O/^t/  -vmK    /'^^^-^ 

273.  It  is  impossible  to  find  an  even  root  of  a  negative 
number;  for  no  number  when  raised  to  an  even  power  can 
produce  a  negative  result  (§  96). 

^/■A'      An  Imaginary  Number  is  an  indicated  even  root  of  a  negative 
number;  as  V— 2,  or  ^—3. 

In  contradistinction,  rational  and  irrational  numbers  (§  248) 
are  called  reed  numbers. 

274.  An  imaginary  number  of  the  form  V— a  is  called  a 
►            piire    imaginary  number,    and    an    expression    of    the    form 

a  +  6V—  1  a  complex  number. 


SURDS  243 

275.  Meaning  of  a  Pure  Imaginary  Number. 

If  Va  is  real  (§  273),  we  define  Va  as  an  expression  such 
that,  when  raised  to  the  second  power,  the  result  is  a  (§  206). 

To  find  what  meaning  to  attach  to  a  pure  imaginary  number, 
we  assume  the  above  principle  to  hold  when  V«  is  imaginary. 

Thus,  V— 2  means  an  expression  such  that,  when  raised  to 
the  second  power,  the  result  is  —  2 ;  that  is,  (V—  2)^  =  —  2. 

In  like  manner,  (V—  1)^  =  —  1 ;  etc. 

OPERATIONS  WITH  IMAGINARY  NUMBERS 

276.  By  §275,  (V^-=-5.  (1) 
Also,  (V5V^'  =  (V5)2(V^2^5(-l)=-5.  (2) 
From  (1)  and  (2),  ( V^'  =  (V5  V^'. 

Whence,  V— 5=V5V— 1. 

Then,  every  imaginary  square  root  can  he  expressed  as  the 
product  of  a  real  number  by  V— 1., 

V—  1  is  called  the  imaginary  unit;  it  is  usually  represented  by  i. 

277.  Addition  and  Subtraction  of  Imaginary  Numbers. 
Pure  imaginary  numbers  may  be  added  and  subtra<;ted  in 

the  same  manner  as  surds. 


1.   Add  V -  4  and  V^^^. 

By  §276,      yr^  +  V^^^  =  2V^l  +  6^ 


2.    Subtract  3 - V-9  from  1  +  V- 16. 

In  adding  or  subtracting  complex  numbers,  we  assume  that  the  rules 
for  adding  or  subtracting  real  numbers  may  be  applied  without  change. 

Then,      I  +  V^^iq  _ (^^  -  \/~^)  =  1  +  ^V^  -  3^+ ^y/^^ 

EXERCISE  119 

Simplify  the  following : 


1.   V^  +  V^^^25.  -i    2.    V^r5+V-45. 


244 


3.    V^r27_V-12. 


ALGEBRA  ,  ^.^  "" 

i        iJ>       ll 

6.    V-64  +  V-100  +  V-121. 


7.   2V-16-5V-49  +  3V-81.     ,- 


XT' 


-4.    V-(a;  +  l)'-V-ar^. 

W    i 

10.  V-a'-2a-l-V-a2  +  2a-l. 

11.  Add  5  +  V^^  to  3  +  V^^16.        T  -r  t'lft 

12.  Add  6-V^=^  to  1-V^49. 

13.  Subtract  2  +  V^^  from  8-V^25. 

14.  Subtract  4-V-81  from  7+V^^^. 

278.  Positive  Integral  Powers  of  V— 1. 
By  §275,  {-V^y^-l, 
Then, 

(V3i)3=(V3T)2x  V^ri  =(_i)xV3T  =  _V3ij 

(V=T)^=(V^)^x(V:rT)2=(_l)x(_i)  =  i; 

(V3i)«=(V3I)4x  V^Ti    =      1  xV^=V^::i;  etc. 

Thus,  the  first  four  positive  integral  powers  of  V— 1  are 
V— 1,  —1,  —  V— 1,  and  1;  and  for  higher  powers  these 
terms  recur  in  the  same  order. 

279.  Multiplication  of  Imaginary  Numbers. 

The  product  of  two  or  more  imaginary  square  roots  can  be 
obtained  by  aid  of  the  principles  of  §§  276  and  278. 

1.   Multiply  V"^  by  V"^. 

By  §276,   V^r2xvCr3  =  V2V:rTxV3v^ZT 

=  V2 \/3(  V'^y  =  V6( -  1)  (§  278)  =  -  V6. 


SURDS 


-^ 

'-%- '' 


245 


2.  Multiply  together  V-9,  V-16,  and  V-25. 

=  60(\/^n:)3  =  60(-V"^)  (§278)  =-60  v/^^. 

3.  Multiply  2  +  5V^=^  by  4-3V^^. 

In  multiplying  complex  numbers,  we  assume  that  the  rules  for  multi- 
plying real  numbers  may  be  applied  without  change. 


2  4-    5V-5 

4_  zv^rs 


8  +  20V-5 


15(-5) 


8  +  14V-  5  +  75  =  83  +  14 V^. 

4.   Expand  (V^^  +  2V^^)'  by  the  rule  of  §  97. 
(  V35  +  2\/^)2  =  (  V^2  +  4\/5  V^Tx  VSV^^  +  4(\/^2 

=  -  6  +  4  vT5(\/^=T)2  +  4(- 3)  =  -  5  -  4  Vl5  - 12  =  -  17  -  4  Vl5. 


EXERCISE  120 


Multiply  the  following 
^1.    V^3  by  V^=^. 

3. 


36  by  -V-25. 


5.  V-14  by  ^-56. 

6.  _V^=^147  by  -V^^^45. 


V-81a^ by -V -121x2.  7^  5_|.4V-1  by  2-3V"=^. 
4.  _V3i5  by  V^^.      8.  6  +  V^3  by  7  +  4V^3. 
9.  3V^^-2V^  by  9V^^  +  6V^. 
10.  SV-T-TV^^^  by  V^^-5V^^. 
V^^^  V^^TF,  and  -V^=^9?. 
V— 6, 


11. 
12. 


2,  V-462,  and 
V^^27,  and 


V-54. 
13.  V-27+V^^^  by  V-"3-V^^. 
Sl4.  2V^^-V^6  by  V^ni  +  4V^. 
15.  V^^^^,  V^=49,  V^=^,  and  V^^^IOO. 


/16.  V-2,  -V-3,  -V^6,   and  -V-10. 


246 


ALGEBRA 


Expand  the  following  by  the  rules  of  §§  97,  98: 


•17.    (5-l-V^l 

^19.   (4V^^  +  3V^)2. 
^^23.   (4V^=^  +  5- 
24.    (8V^=^  +  3- 


20.  (3V-5-2V^^y. 

21.  (7+2V^(7-2V^. 


22.    (V-a4-?>)(V-a-6). 
—  i/)(4v  — ic  —  5v^—  y). 


/  25.   (3V^=34-V-3)2  +  (3V=^-V-3)2. 

Expand  the  following  by  the  rules  of  §  205 : 
^26.    (1-V^^  27.    (2  +  V=^)». 

28.   Expand  (3  V^^  -  V^  -  2  V^3)2  by  the  rule  of  §  204. 


280.   Division  of  Imaginary  Numbers. 


1.   Divide  V-40  by  V-5. 

\/^r40     Vio  V^^     V40 


By  §  276, 


-5 


V5  v^n 


2.   Divide  Vl5  by  V"^. 

Vl5  ^_vT5(-l)^-\/l5(V^ 


V5 


D^ 


=  VS  =  2  \/2. 


278)  =  -  V5  ^^=3  =  -  \/^. 


3.   Reduce  — to  an  equivalent  fraction  having  a 

real  denominator. 

We  multiply  both  terms  of  the  fraction  by  the  denominator  with  the 
sign  between  its  terms  reversed  ;  multiplying  both  terms  by  V3  —  V— 2, 

V3- 


^=    C^-^^)^    (§98) 

V3  +  Vr2    (V3)2_(V3r2)2 

_(V3)2_2V3  V^=^+( 
3-(-2) 

_3-2\/^-2_l-2V^r6 
3  +  2  6 


2}f 


(§97) 


SURDS 


247 


EXERCISE  121 


Divide  the  following : 

^1.  v=n:5  by  v^^. 

^.   -V48  by  V^. 


^3.  V=^  by  -V-8. 


7.  V343-V-63  by 


4.  —-V—Qxy  by  —^2yz. 
^15.  Vl80  by  -V^n^. 
6.  -Vi32  by  -V^=^. 


"^^.  V-288-V300  by  -V-6. 

Eeduce  each_of_ilie  following  to  an  equivalent  fraction  hav- 
ing a  real  denominator : 

3V-3-2 


9. 


V 


10. 


1-V- 
3-V- 


3+V^^ 


11. 


12. 


3V-3+2V-6 

2V^^4-7V^3 
4V^=^-3V^^ 


281.   The  complex  numbers  a  +  ftV—  1  and  a  —  6  V—  1  are 
called  Conjugate. 

We  have       (a-{-b  V"^)  +  (a  -  6  V"^^)  =  2  a. 

Also,  (a  +  b  V"^)  X  (a  -  6  V^^) 

=  a^  -  &2<(  V^^)2  =  a2  _^  62  (§  275). 

Hence,  the  sum  and  product  of  two  conjugate  complex  numbers 
are  real. 


248  ALGEBRA 


XIX.    QUADRATIC  EQUATIONS 

I.   A  Quadratic  Equation  is  an  equation  of  the  second  degree 
(§  83),  with  one  or  more  unknown  numbers. 

A  Pure  Quadratic  Equation  is  a  quadratic  equation  involving 
only  the  square  of  the  unknown  number ;  as,  2  a^  =  5. 

An  Affected  Quadratic  Equation  is  a  quadratic  equation  involv- 
ing both  the  square  and  the  first  power  of  the  unknown  number ; 
as,  2a^-3aj-5  =  0. 

In  §  126,  we  showed  how  to  solve  quadratic  equations  of  the  forms 
ax2  -\-bx  =  0,  ax^  +  c  =  0,  x^-\-ax  +  b  =  0,  and  ax^-\-bx  +  c  =  0, 
when  the  first  members  could  be  resolved  into  factors. 

PURE  QUADRATIC  EQUATIONS 

283.  Let  it  be  required  to  solve  the  equation 

a^  =  4 
Taking  the  square  root  of  each  member,  we  have 

±«=±2; 

for  the  square  root  of  a  number  may  be  either  -f  or  —  (§  208). 

But  the  equations  —x  =  2  and  —  x=  —  2  are  the  same  as 
x  =  —  2  and  x  =  2,  respectively,  with  all  signs  changed. 

We  then  get  all  the  values  of  x  by  equating  the  positive  square 
root  of  the  lirst  member  to  ±  the  square  root  of  the  second. 

284.  A  pure  quadratic  equation  may  be  solved  by  reducing 
it,  if  necessary,  to  the  form  a^  =  a,  and  then  equating  jc  to  ± 
the  square  root  of  a  (§  283). 

1.    Solve  the  equation  3  a;^  _^  7  ^  ^  _^  35^ 

Clearing  of  fractions,  12  a;^  +  28  =  5  tc*  4.  140. 

Transposing,  and  uniting  terms,         .7  x^  =  112,  or  x^  =  16. 
Equating  a;  to  ±  the  square  root  of  16,  x  =  ±  4. 


QUADRATIC   EQUATIONS  249 

2.   Solve  the  equation  7  x^  — 5  =  5  oiy^  — IS. 
Transposing,  and  uniting  terms,  2  a:2  —  _  8,  or  x^  =  —  4. 

Equating  x  to  ±  the  square  root  of  —  4,  x  =  ±  V—  4 

=  ±2\/:ri(§276). 

In  this  case,  both  values  of  x  are  imaginary  (§  274) ;  it  is  impossible  to 
find  a  real  value  of  x  which  will  satisfy  the  given  equation. 


In  solving  fractional  quadratic  equations,  any  solution  which  does  not 
satisfy  the  given  equation  must  be  rejected. 
Thus,  let  it  be  required  to  solve  the  equation 

x2-7  1  1 


x^  +  x-2     x  +  2     x-1 

Multiplying  both  members  by  (x  +  2)  (a;  —  1),  or  x^  +  jc  —  2, 

x^-7  =  x-l-x-2,  or  x2  =  4. 

Extracting  square  roots,  x  =  ±2. 

The  solution  x  =  —  2  does  not  satisfy  by  the  given  equation ;  the  only 
solution  is  X  =  2. 

EXERCISE  122 

Solve  the  following  equations : 

1.    2a;2  +  27  =  7a^-53.  2.    A_15-  =  _?. 

4.af     8a^  3 

3.    5(2x-3)-\-2x(4.x-\-l)=12x-7, 

V  4.    2(3x-5y  +  3(x-\- 10)2 ^ 434^ 

^    '     3       4.x       9       4  a;*  -j^^ 


6.    6-V5ar^-9  =  12.  V?^ 

9.    (2 a; +  7)  (5 a; -6) -24 a;  =  (4 a;- 3) (7  a; +  5) -59. 

10    4^2  +  3     8f2-l^  1  jj        3a  x  +  5h    ^^ 

7  2  14*  *    a;-56     3a  +  105 


250  ALGEBRA 

\J12     3a^  +  7      5a^  4-3^40)^-10 
7  14  35      * 

13.    {x-{-a){x-^b)  +  {x-a){x-b)=x'  +  a^-{-ly^. 


14.   3V^Tl+V3a.'  +  7  =  l. 


15. 


lOa^-3     5x^-i-6     6x^-1 


18  9  9x^-2    ' 

16.    (k  +  l)(k-2)(k-S)-(k-l)(k  +  2)(k-\-3)  =  ^52. 
^17.    2xV^^TS-2x-y^^+2  =  l. 
18     3^!zi^_if^!±3^2aj*  +  12_^ 


0,^  +  5        a^_5        a;4_25 

3a^-l  ^  a;^  +  3         ^q     a 

2aj^-5a;2  +  l      2x^-5  '    '  x-2  x  +  3 


19      a;^  +  3a^-l  ^  a;^  +  3         ^q     x'-x^2     x'j^x-^^j^ 


21. 


Va2  +  ax  +  a^4-Va2-aa;  +  a;2  =  a(l  +  V3).  (^^ 

22    -J^ ^  =  _^!:zlL_.  Z.^-''  0 

•    x  +  3     x-5     x'-2x^l5  /C^f       ^., 

(^       ^"^^ 

AFFECTED  QUADRATIC  EQUATIONS  >   ^f    / Z' 

285.   First  Method  of  Completing  the  Square.  V*     /  ^ 

By  transposing  the  terms  involving  x  to  the  first  member,  and 

all  other  terms  to  the  second,  and  then  dividing  both  members 

by  the  coefficient  of  x^,  any  affected  quadratic  equation  can  be 

reduced  to  the  form  x^-{-px  =  q. 

We  then  add  to  both  members  such  an  expression  as  will 

make  the  first  member  a  trinomial  perfect  square  (§  111) ;  an 

operation  which  is  termed  completing  the  square. 

Ex.     Solve  the  equation  x^  +  Sx  =  4:. 

A  trinomial  is  a  perfect  square  when  its  first  and  third  terms 
are  perfect  squares  and  positive,  and  its  second  term  plus  or 
minus  twice  the  product  of  their  square  roots  (§  111). 

Then,  the  square  root  of  the  third  term  is  equal  to  the  second 
term  divided  by  twice  the  square  root  of  the  first. 


QUADRATIC  EQUATIONS  251 

Hence,  the  square  root  of  the  expression  which  must  be  added 
to  x^  -\-^x  to  make  it  a  perfect  square  is  Sx-i-2x,  or  f. 
Adding  to  both  members  the  square  of  f ,  we  have 

Equating  the  square  root  of  the  first  member  to  ±  the  square 
root  of  the  second  (compare  §  283),  we  have 

Transposing  f,  «  =  —  f-f-for— f  —  |  =  lor—  4. 

We  then  have  the  following  rule : 

Reduce  the  equation  to  tJie  form  oi?  +  px  —  q. 

Complete  the  sq^iare,  by  adding  to  both  members  the  square  of 
one-half  the  coefficient  of  x. 

Equate  the  square  root  of  the  first  member  to  ±  the  square  root 
of  the  second,  and  solve  the  linear  equations  thus  formed. 

286.    1.   Solve  the  equation  3  cc^  —  8  aj  =  —  4. 
Dividing  by  3,  a;2_^^_|, 

which  is  in  the  form  x^  +px  =  q. 

4 

Adding  to  both  members  the  square  of  -,  we  have 

o 


--¥-(ir=-i-f=t' 


Equating  the  square  root  of  the  first  member  to  ±  the  square  root 

Transposing  -|,  x  =  |±|  =  2or|. 

If  the  coefficient  of  oc^  is  negative,  the  sign  of  each  term  must 
be  changed. 

2.   Solve  the  equation  —  9  a^  —  21  a;  =  10. 
Dividing  by  -  9,  x2  +  ^  =  -^- 


252  ALGEBRA 


7 
Adding  to  both  members  the  square  of  -, 

6 


^+'-^+(l)'=-f+*i=' 


3       \6  9       36     36 


7         3 
Extracting  square  roots,  x  +  -  =  ±-' 

6         6 

Then.  .  =  _Z±|  =  _|„._|. 


EXERCISE  123 

Solve  the  following  equations  : 

1.  x^-\-Sx  =  9.               '  8.  3a^  +  Sx  =  -4:, 

2.  x^-4.x  =  60.  /9.  2a^  +  9ic  =  -4. 

3.  x'-dx^-U.                .  10.  4a^+15a;-25  =  0. 

4.  x'+llx  =  -24.,  11.  9ar^  +  14  =  27a;. 

5.  a^-a;  =  20.  12.  18  =  4a?2  _^21x. 

6.  3x2-10x  =  -3.  13.  12-7a;-10a;2  =  0. 

7.  5a^  +  3a;  =  8.  14.  19^  =  -12^-5. 

287.  If  the  coefficient  of  a^  is  a  perfect  square,  it  is  con- 
venient to  complete  the  square  directly  by  the  principle  stated 
in  §  285 ;  that  is,  by  adding  to  both  members  the  square  of  the 
quotient  obtained  by  dividing  the  coefficient  of  x  by  twice  the  square 
root  of  the  coefficient  of  x^. 

1 .   Solve  the  equation  9  a^  —  5  a?  =  4. 

Adding  to  both  members  the  square  of  — - — ,  or  -, 
^  ^  2x3'        6' 


\Q)  36 


169 
36 


C  1  O 

Extracting  square  roots,  3  x  —  -  =  i  —  • 

6  6 

Then,  3cc=-±— =3  or  --,  and  x=l  or  --. 

6     6  3  9 


QUADRATIC   EQUATIONS  253 

If  the  coefficient  of  ^  is  not  a  perfect  square,  it  may  be  made 
so  by  multiplication. 

2.   Solve  the  equation  8  ar^  — 15  a;  =  2. 
Multiplying  each  term  by  2,  16  aj^  —  30  a;  =  4. 

Adding  to  both  members  the  square  of  ,  or  — , 

2x4  4 


16.^-30..(^y  =  4.f  =  f, 


15         17 
Extracting  square  roots,  4  x  — -  —  ±-—. 

4  4 

Then,  4a;  =  — ±  —  =  8  or  -1,  and  a;  =  2  or  -i. 

4       4  2  4 

If  the  coefficient  of  x^  is  negative,  the  sign  of  each  term  must  be 
changed. 

EXERCISE  124 
Solve  the  following  equations : 

1.  4a;2-7x  =  -3.  8.  36 a;^ - 36 or  =  7. 

2.  9a^  +  22a;  =  -8.  9.  Vlx^-^x=\. 

3.  16a^-8a;  =  35.  10.  49 ^^ .f 49 /i  + 10  =  0. 

4.  '^y?^-\^x  =  X  11.  64ar^  +  15  =  64a;. 

5.  3a;2-8a;  =  3.  /12.  12  =  23e-5e2. 

6.  18a^-5a;  =  2.  13.  28a;-32x2_3^0. 

7.  25a;2  +  15a;  =  4.  14.  25 a;  = -50 a;^- 2. 

288.   Second  Method  of  Completing  the  Square. 

Every  affected  quadratic  equation  can  be  reduced  to  the  form 
a'31?  +  6a;  4-  c  =  0,  or  aa^  -(-  6a;  =  —  c. 

Multiplying  both  members  by  4  a,  we  have 

4  aV  -H  4  abx  =  —  4  ac. 

We  complete  the  square  by  adding  to  both  members  the  square 

of  ^-^^^(§287),  or  6. 


254  ALGEBRA 

Then,  4  o V  +  4  ahx  +  52  ^  &2  _  4  ^^^ 


Extracting  square  roots,  2ax-\-'b  =  ±  Vb^  —  4  ac. 
Transposing,  2ax  =  —b±  Vb^  —  4  ac. 


„7.,  —b±  V6  —  4  ac 

Whence,  x  = =^^— —  • 

2a 

We  then  have  the  following  rule : 

Reduce  the  equation  to  the  form  ax^  -{■bx  =  —  c. 
Multiply  both  members  by  four  times  the  coefficient  of  7?,  and 
add  to  each  the  square  of  the  coefficient  of  x  in  the  given  equation. 

The  advantage   of  this   method  over  the  preceding  is  in 
avoiding  fractions  in  conipleting  the  square,  v 

1.  Solve  the  equation   23i^  —  7x  =  —  3, 
Multiplying  both  members  by  4  x  2,  or  8, 

16x^-66x  =  -2i. 

Adding  to  both  members  the  square  of  7, 

16x2  -  56x  +  7^  =  -  24  +  49  =  25. 

Extracting  square  roots,  4  x  —  7  =  ±  5. 

Then,  4  x  =  7  ±  5  =  12  or  2,  and  x  =  3  or  i. 

2' 

If  the  coefficient  of  x  in  the  given  equation  is  even,  fractions 
may  be  avoided,  and  the  rule  modified,  as  follows : 

Multiply  both  members  by  the  coefficient  of  x^,  and  add  to  each 
the  square  of  half  the  coefficient  of  x  in  the  given  equatio7i. 

2.  Solve  the  equation  15  a^^  _|_  28  a;  =  32. 

Multiplying  both  members  by  15,  and  adding  to  each  the  square  of  14, 

152x2  +  15  (28  X)  +  142  =  480  +  196  =  676. 

Extracting  square  roots,  15x  +  14  =  ±  26. 

Then,  15  x  =  -  14  ±  26  =  12  or  -  40,  and  x  =  -  or  -  §• 

5  3 

The  method  of  completing  the  square  exemplified  in  the  present  section 

is  called  the  Hindoo  Method. 


QUADRATIC   EQUATIONS  255 


EXERCISE  125 

Solve  the  following  (equations  : 

^  1.  x\7x=18.  9.  12x'-llx  =  -2. 

2.   3x'-2x  =  4:0.  10.  6x'-13x  =  -6. 

-3.   4.x'-Sx  =  10.  11.  2r2-15r  +  25  =  0. 

4.  4.x^-Sx  =  4.5.  12.  15a:2  4-26aj  +  7  =  0. 

5.  8a;2  +  2a;  =  3.                       13.  5a;2+ 48  =  -32a;. 
^6.   9a^  +  18a;  =  -8.                 14.  13a;==  lOcc^-S. 

7.   9aj2  +  4a;  =  5.  15.  3  =  6 a;^  + 17 a;. 

^>i8.   7g2  +  20g  =  -12.  16.  27i»-9 -8^^2  =  0. 

289.   Solution  of  Affected  Quadratic  Equations  by  Formula. 

It  follows  from  §  288  that,  if  ax^  -\-bx-\-c  =  0,  S^^w^mO 

then  ^^-6±V6--4ac.      ^t*^.,^^  (1) 

Z  ft 

This  result  may  be  used  as  a  formula  for  the  solution  of  any 
affected  quadratic  equation  in  the  form  aaf  -\-bx  +  c  =  0. 

1.    Solve  the  equation  2x^-\-5x  —  lS  =  0. 

Here,  .a  =  2,  6  =  5,  and  c  =  —  18  ;  substituting  in  (1), 


_5j:V25  +  144^-5  ±13^^  ^^      9 
4  4  2 

2.   Solve  the  equation  —  5  o?^  + 14  a;  +  3  =  0. 
Here,  a  =  —  5,  6  =  14,  c  =  3  ;  substituting  in  (1), 


-  14  j-  V196  +  60  ^  -  14  zfc  16  ^     1 
-10  ~       -10  5 

3.   Solve  the  equation  110  a^  -  21  a;  =  - 1. 
Here,  a  =  110,  &  =  -  21,  c  =  1 ;  then, 


21  i  V441  -  440     21  ±  1       1    «^   1 
a;  =:  — £= = =  —  or  — • 

220  220        10        11     • 

Particular  attention  must  be  paid  to  the  signs  of  the  coefficients  in 
making  the  substitution. 


256  ALGEBRA 

EXERCISE  126 

Solve  the  following  by  formula : 

1.  aj2_i2a;_f-32  =  0.    :  8.   40- 17 a;-5a^  =  0. 

2.  a^  +  7a;-30  =  0.    ^j-'^      9.   36p'-\-S6p  =  -5. 

3.  2a;2-3a;-20  =  0.  H^  .^-  10.   SOic^  +  l  =  -17a;. 

4.  3x^-x-4.  =  0.  I     -/  *'ll.    -19a;  =  8aj2_^6. 

5.  4a!2_5^_2i  =  o^  3^.^    12.   15 +  22aj- 48x2  =  0. 

6.  20a^  +  aj-l  =  0.  ^^.  .  <^    13.   l^x'  +  2Qx^-S. 

7.  9a.-2-18a;  +  8  =  0.  v^    ^   14.   37a;  =  6a^  +  6. 

3  '  J 
EXERCISE  127 

The  following  miscellaneous  equations  may  be  solved  by  either  of  the 
preceding  methods,  preference  being  given  to  the  one  best  adapted  to  the 
example  under  consideration. 

In  solving  any  fractional  equation,  we  reject  any  solution  which  does 
not  satisfy  the  given  equation.     (Compare  last  example,  §  284.) 

10  3    ^ 13  ^  1 

3*  '6a;     ^x^     18* 

5  .    A  +  i^  =  _15 

12a;*  *    U      3  6* 

5.  (3a;  +  2)(2a;-3)  =  (4a;-l)2-14. 

6.  a;(5a;  +  22)+35  =  (2a;  +  5)l 

7.  (a;  +  4)(2a;-l)  +  (2a;-l)(3a;  +  2)  =  (3a;  +  2)(4a;-l)-49. 

8.  ^1:^-^  =  1.  10.    _A_4._A_  =  3. 

6~  X     8  — a; 

6a;  +  5_4a;  +  4 
4a;-3~  a;-3  * 

15        3a;         4  —  ^x  _     5 

14.   (a; -4)3 -(a; +  3)3  =  —217.        *    4-5a;         3a;     ~~6* 


1. 

a;2 
2 

7a; 
6 

2. 

7 
6a;2 

1 

2 

24        24  _  ^ 

a;-2       X 

10. 

d  +  3     (^  +  4     3 
d-2         d         2 

11. 

(a;  +  l)(a;  +  3)  =  12 

>  +  (^ 

+  7)V2. 

V5a;2-3a;-41  = 

3a;- 

-7.    _ 

QUADRATIC   EQUATIONS  257 

16.  V5a;  +  ll=V3x  +  l  +  2.  a;-2     a;  +  4^     7 

17.  V754-8- V5s-4  =  2.  *    a;  +  5     a;-3~     3 


19.    V8  a;3_35  a;2+55  a;  -  57  =  2  a;  -  3. 

20.   3V^=1 4^  =  4.    22.   28(3.  +  10)_        25  , 

^x  —  1  8  0^^  —  27        a;(2a;  — 3) 

21     2a;  +  l      3a;-2^17       03     _J^ 1       ^14 

•    3a;-2     2a;  +  l      4'  *    x'-Sx     x'-^-Ax     ISa;^* 

24.    ^+       ^^  1^ 


a;-2     24(a;  +  2)      a^-4 

25  5  7      ^8^^-13i;-64 

*  2v-f3     3?;-4        6v'  +  v-12  ' 

26  1^— i__  lU  3/^—1—- i^ 

*  3V4a:-l      2/       V3a;  +  1      3j 

27.    2V3^T4  +  3V3^+7=        ^ 


28. 


V3a;  +  4 
2a^-4a;-3^a^-4a;  +  2 
2x'-2x-\-S     x^-3x-[-2 


29.    -1— --_i_-  =  l+     ^ 


a^-4     3(a;  +  2)  2 -a; 

30.   J^Ii+J^±5  =  ^ 


31.    V2^  +  2V2a;  +  5  =  2V6a;  +  4. 


32.    V8a;  +  7  =  V4a;  +  3  +  V2a;  +  2. 


33.    V2x24-7a;  +  7  =  6-V2a^-9ic-l. 


34.  V6-5aj  +  V2-7a;  =  Vl2  +  6a;. 

35.  ^±^+^±^^.^±3  =  3. 
ic  —  1     x  —  2     x  —  3 

(Compare  Ex.  1,  §  167.) 

Q/j     X  —  2     x-\-2     x—2         -I 


(B  — 4     a;H-3     a;  — 6 


258  ALGEBRA 

37. 


38. 


"^    1 

x-1 

x'^-x- 
x'-x 

1 

x~l  ' 

X 

X 

X 

x'  +  2x 
x'  +  ^x 

-2 

x^2 

X  +  3 

+  6* 

07  +  4 

X 

,.-5 

x 

on       ^-f-^  _      i^      _|_  JO  —  iJ  _      M      ^ 

X  X+^  X  X—5 

(First  combine  the  first  two  fractions,  and  then  the  last  two.) 

290.   Solution  of  Literal  Affected  Quadratic  Equations. 
For  the  solution  of  literal  affected  quadratic  equations,  the 
methods  of  §  288  are  usually  most  convenient. 

1.    Solve  the  equation  x^-{-ax—bx  —  ab  =  0. 

We  may  write  the  equation  x^  -{-  (a  —  b)x  =  ab. 
Multiplying  both  members  by  4  times  the  coefficient  of  x^^ 

4a;2  +  4(a-  b)x  =  4ab. 

Adding  to  both  members  the  square  of  a  —  b, 

4  ic2  +  4(a  -  b)x  +  (a  -  6)2  =  4  a&  +  a2  _  2  a6  +  b'^ 

=  a2  +  2  a&  +  62. 

Extracting  square  root,  2  x  +  (^a  —  b)  =  ±(a  +  b). 

Or,  2x  =  -{a-b)±{a  +  b). 

Then,      *  2x  =  -a  +  b  +  a-\-b  =  2b, 

or  2x  =  -a  +  b-a-b  =  -2a.' 

Whence,  a;  =  6  or  -a. 

If  several  terms  contain  the  same  power  of  x,  the  coefficient  of  that 
power  should  be  enclosed  in  parentheses,  as  shown  in  Ex.  1. 


The  above  equation  can  be  solved  more  easily  by  the  method  of  §  126  ; 
thus,  by  §  108,  the  equation  may  be  written 

(a;  +  a)  (x  -  6)  =  0.  w 

Then,  x  +  a  =  0,  orx=— a; 

and  «  —  6  =  0,  or  a;  =  6. 


QUADRATIC   EQUATIONS  259 

Several  equations  in  Exercise  128  may  be  solved  most  easily  by  the 
method  of  §  126. 

2.    Solve  the  equation  (m  —  l)x^  —  2  m^x  =:  —  4:m^. 

Multiplying  both  members  by  w  —  1,  and  adding  to  both  the  square 
of  m2, 

(m  -  1)%2  —  2  TO2(m  -  l)x  +  w*  =  —  4  m^(m  -  1)  +  m* 

=  m*  —  4  m^  +  4  m^. 

Extracting  square  root,  (m  —  l)x  —  m^  =  ±  (m^  —  2m). 

Then,        (m  —  l)aj  =  m^  +  m^  —  2  w  or  m^  —  m^  -f  2  m 

=  2m(TO  —  1)  or  2  m. 

Whence,  x  =  2  m  or  — ^. 

TO  —  1 

In  solving  any  fractional  equation,  we  reject  any  solution  which  does 
not  satisfy  the  given  equation. 

EXERCISE  128 

Solve  the  following  equations : 
X  1.    a^  4-  2  mx  =  1  —  m^.  /  4.   x^  -{•  nx -i- x  =  —  n. 

2.    x^  —  2ax  =  —  6a-\-9.  ^5.    ay^  —  m^nx -\- mn^x  =  m^n^. 

^.   x^-\-(a  —  b)x  =  ab.  ,6.   a^  — 4aa;  — 10a;  =  — 40  a. 

/L   6a^  +  4aaj-15  6a;  =  10a6.  ^ 

y8.    amx^  —  anx  —  bmx -{- bn  =  0.      J$k    ♦         y 

.    Va+a;  — •v2«= — /lO. M^^     ,     =-«  ^*f 

.        Va+x  2a;  +  a     ^i-4a     3 

yii.  (a-\-xy-\-{b-xy  =  (a  +  bf.       .  i^f,  /  1 


yi2.  V(a  +  26)x-2a6  =  a;-46. 

il3.  (a2-a-2)a;2-(5a-l)ic  =  -6.. 

y^4.  a^  —  (m  — p)  a;  +  (m  —  n)  (ri  — p)  =  0. 

^15.  (a  +  6)a;2+(3a-te&)a;  =  -2a. 

/16.  (b  +  c)x^-{a-{'C)x  =  h-a, 


260  ALGEBRA 

16  a 


^j^7.    VaJ  +  a  +  2 V^  +  6a 


^  ■\Jx  4-  a 


18.    Va;  — a,  +  v2a;  +  3a  =  V5a. 


19. 
20. 


X 


a  +  &^2(a^  +  &'). 
a  4-  6         ic  a^—lP' 

1         _^1      1      1. 

a—b—x     a      b     x 


21.    ^^ + ? =  2. 

(c  +  a  — c     a;  +  6  — c 

22    ^^  —  ^^ _L  '^^  +  n  ^  10^ 

3a7H-n       2x—  Sn      3 

23.    a=-'c2(l  +  a?)'  -  bH%l  -  xf  =  0. 
y24.   ^!^  =  _i^.  25.    2a^  +  l  ^2n4-l, 


26.    Vma;  +  V(m  —  n)x  +  mn  =  2m. 


V  27. 


x  —  a     x-{-  a     a^  —  5a^ 


x-\-a     x  —  a       x^  —  a^ 


^^^^y^^t>'a.^     CV| 


V  28     ^'  +  ^        3 a;  —  2  g _      21  aa;  —  4  g^         ^  I'iC  -y< 

2x  —  3a       3a;  +  g       6a;2  — 7  ga;  — 3  g^        1— 't'V 
^f^9.    (g-6-f 2c)a^-(2g  +  6  +  c)a;  =  -a-26  +  c.'^   ' 

30.   -J_  +  lt-^  +  i  =  0. 
^   l^g     a    M-\-b      b 


-a;      g      g  +  o 


«|  PROBLEMS   IN  PHYSICS 

owing  equations  occur  in  the  study  of  physics. 
Solve  in  the  first  six  equations  for  the  number  which  appears  to  the 
second  power. 

1.   S.^\gt\  3.   F='^.  ^-  •^=^- 


QUADRATIC  EQUATIONS  261 

7.  Solve  the  following  equation  for  a ;  ^  =  7r\/-- 

8.  Solve  the  following  equation  for  t;  S  =  Vot  -{-^  gt\ 

9.  Solve  the  following  equation  for  s ;   V=^2gs. 

10.  In  problem  1  solve  for  g ;  in  problem  7  solve  for  I ;  in 
problem  2  solve  for  m ;  in  problem  4  solve  for  M ;  in  problem 
6  solve  for  Z. 

PROBLEMS  INVOLVING  QUADRATIC  EQUATIONS  WITH 
ONE  UNKNOWN  NUMBER 

291.  In  solving  problems  which  involve  quadratic  equations, 
there  will  usually  be  two  values  of  the  unknown  number;  only 
those  values  should  be  retained  which  satisfy  the  conditions  of 
the  problem. 

1.  A  man  sold  a  watch  for  $21,  and  lost  as  many  per  cent 
as  the  watch  cost  dollars.     Find  the  cost  of  the  watch. 


Let  X  =  number  of  dollars  the  watch  cost. 

Then,  x  =  the  per  cent  of  loss, 


r 

and  X  X  -^,  or  —  =  number  of  dollars  lost. 

100         100 

By  the  conditions,         -^^  =  ic  —  21. 
100 

Solving,  X  =  30  or  70.  ^ 

Then,  the  cost  of  the  watch  was  either  $  30  or  $  70  ;  for  either  of  these 
answers  satisfies  the  conditions  of  the  problem. 

2.  A  farmer  bought  some  sheep  for  $  72.  "^f  h^  had  bought 
6  more  for  the  same  money,  they  would  have  ddlrt  him  $  1 
apiece  less.     How  many  did  he  buy  ? 

Let  n  =  number  bought. 

72 
Then,  —  =  number  of  dollars  paid  for  one, 

n 

72 
and =  number  of  dollars  paid  for^Oine  if  there 


^  "^  had  been  6  more. 


262  ALGEBRA 

72         72 

By  the  conditions,  —  = h  1. 

n      n  +  6 

Solving,  w  =  18  or  —  24. 

Only  the  positive  value  is  admissible,  for  the  negative  value  does  not 
satisfy  the  conditions  of  the  problem. 

Therefore,  the  number  of  sheep  was  18. 

If,  in  the  enunciation  of  the  problem,  the  words  "6  more"  had  been 
changed  to  "6  fewer,"  and  "$1  apiece  less"  to  "|1  apiece  more,"  we 
should  have  found  the  answer  24. 

3.  If  3  times  the  square  of  the  number  of  trees  in  an  orchard 
be  increased  by  14,  the  result  equals  23  times  the  number ;  find 
the  number. 

Let  X  =  number  of  trees. 

By  the  conditions,  S  x"^  +  U  =  23  x. 

Solving,  a;  =  7  or  ^. 

Only  the  first  value  of  x  is  admissible,  for  the  fractional  value  does  not 
satisfy  the  conditions  of  the  problem. 
Then,  the  number  of  trees  is  7. 

4.  If  the  square  of  the  number  of  dollars  in  a  man's  assets 
equals  5  times  the  number  increased  by  150,  find  the  number. 

Let  X  =  number  of  dollars  in  his  assets. 

By  the  conditions,        cc^  =  5  x  +  150. 

Solving,  X  =  15  or  —  10. 

This  means  that  he  has  assets  of  $  15,  or  liabilities  of  $  10. 

EXERCISE  129 

1.  What  number  added  to  its  reciprocal  gives  2^? 

2.  Divide  the  number  24  into  two  parts  such  that  twice  the 
square  of  the  greater  shall  exceed  o  times  the  square  of  the 
less  by  45. 

3.  Find  three  consecutive  numbers  such  that  the  sum  of 
their  squares  shall  be  434. 


QUADRATIC  EQUATIONS  263 

4.  Find  two  numbers  whose  difference  is  7,  and  the  differ- 
ence of  their  cubes  721. 

5.  Find  five  consecutive  numbers  such  that  the  quotient  of 
the  first  by  the  second,  added  to  the  quotient  of  the  fifth  by 
the  fourth,  shall  equal  f |. 

6.  Find  four  consecutive  numbers  such  that  if  the  sum  of 
the  squares  of  the  second  and  fourth  be  divided  by  the  sum 
of  the  squares  of  the  first  and  third,  the  quotient  shall  be  i-§. 

7.  The  area  of  a  certain  square  field  exceeds  that  of  another 
square  field  by  1008  square  yards,  and  the  perimeter  of  the 
greater  exceeds  one-half  that  of  the  smaller  by  120  yards. 
Find  the  side  of  each  field. 

8.  A  fast  train  runs  8  miles  an  hour  faster  than  a  slow 
train,  and  takes  3  fewer  hours  to  travel  288  miles.  Find  the 
rates  of  the  trains. 

9.  The  perimeter  of  a  rectangular  field  is  180  feet,  and  its 
area  1800  square  feet.     Find  its  dimensions. 

\/-  10.  A  merchant  sold  goods  for  $22.75,  and  lost  as  many 
per  cent  as  the  goods  cost  dollars.     What  was  the  cost  ? 

u-  11.  A  merchant  sold  two  pieces  of  cloth  of  different  quality 
for  $  105,  the  poorer  containing  28  yards.  He  received  for  the 
finer  as  many  dollars  a  yard  as  there  were  yards  in  the  piece ; 
and  7  yards  of  the  poorer  sold  for  as  much  as  2  yards  of  the 
finer.     Find  the  value  of  each  piece. 

l^-  12.  A  merchant  sold  goods  for  %  65.25,  and  gained  as  many 
per  cent  as  the  goods  cost  dollars.     What  was  the  cost  ? 

13.  A  has  five-fourths  as  much  money  as  B.  After  giving  A 
%  6,  B's  money  is  equal  to  A's  multiplied  by  a  fraction  whose 
numerator  is  15,  and  whose  denominator  is  the  number  of  dol- 
lars A  had  at  first.     How  much  had  each  at  first  ? 

^'  14.  A  and  B  set  out  at  the  same  time  from  places  247  miles 
apart,  and  travel  towards  each  other.  A's  rate  is  9  miles  an 
hour;  and  B's  rate  in  miles  an  hour  is  less  by  3  than  the 
number  of  hours  at  the  end  of  which  they  meet.    Find  B's  rate. 


264  ALGEBRA 

15.  A  man  buys  a  certain  number  of  shares  of  stock,  pay- 
ing for  each  as  many  dollars  as  he  buys  shares.  After  the 
price  has  advanced  one-fifth  as  many  dollars  per  share  as  he 
has  shares,  he  sells,  and  gains  $  980.  How  many  shares  did 
he  buy  ? 

16.  The  two  digits  of  a  number  differ  by  1;  and  if  the 
square  of  the  number  be  added  to  the  square  of  the  given 
number  with  its  digits  reversed,  the  sum  is  585.  Find  the 
number. 

17.  A  gives  $  112,  in  equal  amounts,  to  a  certain  number  of 
persons.  B  gives  the  same  sum,  in  equal  amounts,  to  14  more 
persons,  and  gives  to  each  $4  less  than  A.  How  much  does 
A  give  to  each  person  ? 

18.  The  telegraph  poles  along  a  certain  road  are  at  equal 
intervals.  If  the  intervals  between  the  poles  were  increased  by 
22  feet,  there  would  be  8  fewer  in  a  mile.  How  many  are 
there  in  a  mile  ? 

19.  A  merchant  bought  a  cask  of  wine  for  $  48.  Having  lost 
4  gallons  by  leakage,  he  sells  the  remainder  at  $  2  a  gallon 
above  cost,  and  makes  a  profit  of  25%  on  his  entire  outlay. 
How  many  gallons  did  the  cask  contain  ? 

20.  The  men  in  a  regiment  can  be  arranged  in  a  column 
twice  as  long  as  it  is  wide.  If  their  number  were  less  by  224, 
they  could  be  arranged  in  a  hollow  square  4  deep,  having  in 
each  outer  side  of  the  square  as  many  men  as  there  were  in 
the  length  of  the  column.     Find  the  number  of  men. 

21.  The  denominator  of  a  fraction  exceeds  twice  the 
numerator  by  2,  and  the  d,ifference  between  the  fraction  and  its 
reciprocal  is  |j.     Find  the  fraction. 

22.  A  man  started  to  walk  3  miles,  intending  to  arrive  at  a 
certain  time.  After  walking  a  mile,  he  was  detained  10  min- 
utes, and  was  in  consequence  obliged  to  walk  the  rest  of  the 
way  a  mile  an  hour  faster.     Find  his  original  speed. 


QUADRATIC   EQUATIONS  265 

23.  A  regimentj  in  solid  square,  has  24  fewer  men  in  front 
than  v/hen  in  a  hollow  square  6  deep.  How  many  men  are 
there  in  the  regiment  ? 

24.  A  rectangular  field  is  surrounded  by  a  fence  160  feet 
long.  The  cost  of  this  fence,  at  96  cents  a  foot,  was  one-tenth 
as  many  dollars  as  there  are  square  feet  in  the  area  of  the 
field.     Find  the  dimensions  of  the  field. 

25.  A  tank  can  be  filled  by  one  pipe  in  4  hours  less  time 
than  by  another ;  and  if  the  pipes  are  open  together  li  hours, 
the  tank  is  filled.  In  how  many  hours  can  each  pipe  alone  fill 
it  ?     Interpret  the  negative  answer. 

26.  A  crew  can  row  down  stream  18  miles,  and  back  again, 
in  7|-  hours.  Their  rate  up  stream  is  1^  miles  an  hour  less 
than  the  rate  of  the  stream.  Find  the  rate  of  the  stream,  and 
of  the  crew  in  still  water. 

27.  A  man  put  $  5000  into  a  savings-bank  paying  a  certain 
rate  of  interest.  At  the  end  of  a  year,  he  withdrew  $  375,  leav- 
ing the  remainder  at  interest.  At  the  end  of  another  year,  the 
amount  due  him  was  $  4968.     Find  the  rate  of  interest. 

28.  A  square  garden  has  a  square  plot  of  grass  at  the  cen- 
tre, surrounded  by  a  path  4  feet  in  width.  The  area  of  the 
garden  outside  the  path  exceeds  by  768  square  feet  the  area  of 
the  path ;  and  the  side  of  the  garden  is  less  by  16  feet  than 
three  times  the  side  of  the  plot.  Find  the  dimensions  of  the 
garden. 

29.  A  merchant  has  a  cask  full  of  wine.  He  draws  out  6 
gallons,  and  fills  the  cask  with  water.  Again  he  draws  out  6 
gallons,  and  fills  the  cask  with  water.  There  are  now  25  gal- 
lons of  pure  wine  in  the  cask.  How  many  gallons  does  the 
cask  hold  ? 

30.  A  and  B  sell  a  quantity  of  corn  for  $  22,  A  selling  10 
bushels  more  than  B.  If  A  had  sold  as  many  bushels  as  13 
did,  he  would  have  received  $  8 ;  while  if  B  had  sold  as  many 
bushels  as  A  did,  he  would  have  received  ^  15.  How  many 
bushels  did  each  sell,  and  at  what  price  ? 


266  •         ALGEBRA 

31.  Two  men  are  employed  to  do  a  certain  piece  of  work. 
The  first  receives  $  48 ;  and  the  second,  who  works  6  fewer 
days,  receives  $  27.  If  the  second  had  worked  all  the  time,  and 
the  first  6  fewer  days,  they  would  have  received  equal  amounts. 
How  many  days  did  each  work,  and  at  what  wages  ? 

32.  A  carriage-wheel,  15  feet  in  circumference,  revolves  in 
a  certain  number  of  seconds.  If  it  revolved  in  a  time  longer 
by  one  second,  the  carriage  would  travel  14400  fewer  feet  in 
an  hour.     In  how  many  seconds  does  it  revolve  ? 

PROBLEMS  IN  PHYSICS 

1.  When  a  body  falls  from  rest  from  any  point  above  the 
earth's  surface,  the  distance,  S,  which  it  traverses  in  any  num- 
ber of  seconds,  t,  is  found  to  be  given  by  the  equation 

in  which  g  represents  the  velocity  which  the  body  acquires 
in  one  second.  The  value  of  g  is  32.15  feet,  or  980  centi- 
meters. 

A  stone  fell  from  a  balloon  a  mile  high ;  how  much  time 
elapsed  before  it  reached  the  earth  ?  • 

2.  If  a  body  is  thrown  downward  with  an  initial  velocity,  Vq, 
then  the  space  it  passes  over  in  t  seconds  is  found  to  be  given 
by  the  equation 

S  =  Vot-{-igf. 

If  the  stone  mentioned  in  Problem  1  had  been  thrown  down 
from  the  balloon  with  a  velocity  of  40  feet  per  second,  how 
many  seconds  would  have  elapsed  before  it  reached  the  earth  ? 

3.  In  the  equation  ^  =  7r\/-,  t  represents  the  time  required 

by  a  pendulum  to  make  one  vibration,  I  represents  the  length 
of  the  pendulum,  and  g  is  the  same  as  in  Problem  1.  Find  the 
length  of  a  pendulum  which  beats  seconds. 

4.  If  a  pendulum  which  beats  seconds  is  found  to  be  99.3 
centimeters  long,  find  from  the  above  equation  the  value  of  g. 


QUADRATIC   EQUATIONS  267 

5.  In  the  equation  F= — ~,  M  and  m  represent  the  masses 

(I 

of  any  two  attracting  bodies,  as,  for  instance,  the  earth  and  the 
moon,  d  represents  the  distance  between  these  bodies,  and  F  the 
force  with  which  they  attract  each  other. 

If  the  moon  had  twice  its  present  mass  and  were  twice  as  far 
from  the  earth  as  at  present,  how  much  greater  or  less  would 
the  force  of  the  earth's  attraction  be  upon  it  than  at  present  ? 

6.  In  the  equation  E  =  \  mv^,  E  represents  the  energy  of  a 
moving  body,  the  mass  of  which  is  m  and  the  velocity  is  v. 
Compare  the  energies  of  two  bodies,  one  of  which  has  twice  the 
mass  and  twice  the  velocity  of  the  other. 

7.  When  a  bullet  is  shot  upward  with  a  velocity,  v,  the  height, 
8,  to  which  it  rises  is  given  by  the  equation  ^ 

Find  with  what  velocity  a  body  must  be  thrown  upward  to 
rise  to  the  height  of  the  Washington  Monument  (b6b  feet). 
(See  Problem  1.) 


268  ALGEBRA 


XX.     EQUATIONS    SOLVED    LIKE 
QUADRATICS 

292.  Equations  in  the  Quadratic  Form. 

An  equation  is  said  to  be  in  the  quadratic  form  when  it  is 
expressed  in  three  terms,  two  of  which  contain  the  unknown 
number,  and  the  expoyient  of  the  unknown  7iumber  in  one  of 
these  terms  is  twice  its' exponent  in  the  other;  as, 

a;«-6i^=16;  a^-{-x^-72  =  0;  etc. 

293.  Ec^uations  in  the  quadratic  form  may  be  readily  solved 
by  the  rules  for  quadratics. 

1.  Solve  the  equation  x^  —  6a:^  =  16. 
Completing  the  square  by  the  rule  of  §  285, 

x6  _  6  a;3  +  9  =  16  +  9  =  25. 

Extracting  square  roots,      x^  —  3  =  ±  6. 

Then,  '  a;^  z=  3  ±  5  =  8  or  -  2. 

Extracting  cube  roots,  x  =  2  or  —  \/2. 

There  are  also  four  imaginary  roots,  which  may  be  found  by  the 
method  of  §301. 

2.  Solve  the  equation  2x-\-  SVx  =  27. 

Since  Vx  is  the  same  as  x^,  this  is  in  the  quadratic  form. 
Multiplying  by  8,  and  adding  3^  to  both  members  (§  288), 

16  X  +  24\/S  +  9  =  216  +  9  =  225. 

Extracting  square  roots,    4  Vx  +  3  =  ±  15. 

Then,  4  Vx  =  -  3  ±  15  =  12  or  -  18. 

Whence,  Vx  =  3  or  -  -,  and  x  =  9  or  — . 


3.    Solve  the  equation  16  x'^  —  22  x~^  =  3. 


EQUATIONS  SOLVED  LIKE  QUADRATICS         269 

Multiplying  by  16,  and  adding  ll"-^  to  both  members, 

162  x"^  _  16  X  22  x~*  +  112  ^  48  +  121  =  169. 

_3 

Extracting  square  roots,      16  x  ^  —  11  ==  i  13. 

Then,  16  x~^  =  11  ±  13  =  24  or  -  2,  and  x~^  =  |  ^^  ~  |* 

Extracting  cube  roots,  a:"^  ~  ( 9  )  ^  °^  ~  i' 

Raising  to  the  fourth  power,  x-^  "  (9  )^  ^^  i«* 

.  l  =  (?)tori,a„d.  =  (|)iorl6. 

P 

To  solve  an  equation  of  the  form  aj*  =  a,  first  extract  the  root  corre- 
sponding to  the  numerator  of  the  fractional  exponent,  and  afterwards 
raise  to  the  power  corresponding  to  the  denominator ;  careful  attention 
must  be  given  to  algebraic  signs  ;  see  §§  96  and  209. 

EXERCISE  130 

Solve  the  following  equations : 

1.  a;4  -  29  aj2  =  - 100.  5.    a^-^  -  63  a;"^  -  64  =  0. 

2.  x'^  +  19x-'=216.  ^6.   3x-'-\-Ux-'  =  5.     ' 

3.  aj^-10a;^  +  9  =  0.  7.   5 a;"^  +  7 ic"^  =  - 2.  ^-, 

4.  a;^+33a^^  =  -32.  8.   4v'^  +  6  =  ll^^. 

ulO.   9(aj-^  +  lf=(a;-3-4)2  +  llflj-3_5.    ^  )(--^4- /ir)( 

11.  6h-2  =  llVTi.  ,17.   32V^_33  = ^. 

12.  x-^'-  +  2Ux-^  =  -243,  ^^ 

3n  3n  /  18.      161  0^  +  5  =  -  32  OJ^^ 

13.  Sx^-Ax'  =16. 

14.  2.-«-35.-  +  48  =  0.      -    19-  |-308  =  64a.t. 

15.  27x3  +  46  =  g.  ^^     V^  +  V^^^_^V6. 

16.  16a;»-33»*-243  =  0.  *    Va      Va?      V6      Va 


\  - 


270  ALGEBRA 


^   -♦ 


21.   V6+V^-j-V4-Va;  = 


12 


V4-Va; 


/^22.    V3Va;  +  l-f-VV^-4  =  V4Va;+5. 

294.  An  equation  may  sometimes  be  solved  with  reference 
to  an  expression,  by  regarding  it  as  a  single  letter. 

1 .    Solve  the  equation  {x  -  5)^  -  3(a;  -  5)^  =  40. 
Multiplying  by  4,  and  adding  3^  to  both  members, 

4(0;  -  5)3  -  12(x  -  5)  2  +  32  =  160  +  9  =  169. 
Extracting  square  roots,    2(x  —  5)^  —  3  =  ±  13. 


Then, 

2Ca;-5)2  =  3dbl3  =  16  or 

-10. 

Whence, 

{x-  6)^  =  8  or  -  5. 

Extracting  cube  roots, 

(x  _  5)^  =  2  or  -  VI, 

Squaring, 

X  -  5  =  4  or  v'25. 

Whence, 

a;  =  9  or  5  +  v'25. 

Certain  equations  of  the  fourth  degree  may  be  solved  by,  the. 
rules  for  quadratics. 

2.    Solve  the  equation  a;*  +  12a^  +  34aj2- 12  a;- 35  =  0. 
The  equation  may  be  written 

(x*  +  12  x3  +  36  x2)  -  2  x2  -  12  X  =  35. 
Or,  (x2  +  6x)2-2(x2  +  6x)=35. 

Completing  the  square,  (x2  +  6  x)2  -  2(x2  +  6  x)  +  1  =  36. 
Extracting  square  roots,  (x2  +  6  x)  -  1  =  ±  6. 

Then,  x2  +  6  x  =  7  or  -  6. 

Completing  the  square,  x^  +  6  x  +  9  =  16  or  4. 

Extracting  square  roots,  x  +  3=±4or  ±2. 

Then,  x  =  -  3  ±  4  or  -  3  ±  2  =  1,  -  7,  -  1,  or  -  5. 

In  solving  equations  like  the  above,  the  first  step  is  to  complete  the 
square  with  reference  to  the  x*  and  x^  terms  ;  by  §  287,  the  third  term  of 
the  square  is  the  square  of  the  quotient  obtained  by  dividing  the  x^  term 
by  twice  the  square  root  of  the  x*  term. 


EQUATIONS   SOLVED   LIKE   QUADRATICS  271 


3.    Solve  the  equation  af  —  6x-\-  5 Va;^  —  6  a;  +  20  =  46. 
Adding  20  to  both  members, 


(x2  -  6  x  +  20)  +  5Va;2  -  6  x  +  20  =  66. 

Completing  the  square, 

(a;2  _  6 X  +  20)+ 5 Vx2  -  6 X  +  20  +^  =  6Q+^=z^. 

4  4        4 

17 


Extracting  square  roots,     Vx^  —  6x  +  20  +  -  =  ± 


Then,  Vx'-^  -  6  x  +  20  =  6  or  -  11. 

Squaring,                 ^  x2  -  6  x  +  20  =  36  or  121. 

Completing  the  square,  x^  —  6  x  +  9  =  25  or  110. 

Extracting  square  roots,  x  —  S=±5  or  ±  VllO. 

Then,  x  =  8,  -  2,  or  3  ±  vHO. 

In  solving  equations  of  the  above  form,  add  such  an  expression  to  both 
members  that  the  expression  without  the  radical  sign  in  the  first  member 
may  be  the  same  as  that  within,  or  some  multiple  of  it. 

4.   Solve  the  equation  — h  — — -  =  -• 

ar  —  xx^  —  S2 

^2  o 

Representing  — by  y,  the  equation  becomes 

x^  —  X 

y  +  -  =  l,  or  2?/2  +  2  =  5y. 
y     2 

Solving  this,  w  =  -  or  2  :  that  is,  5!jZ-§  =  1  or  2. 
^        '  ^      2  '  '  x2  -  X      2 

Taking  first  value,  2  x^  —  6  =  x^  —  x,  or  x^  +  x  =  6. 

Solving,  X  =  2  or  —  3. 

Taking  second  value,  x^  — ,3  =  2  x^  —  2  x,  or  —3^+2  x=3. 

Solving,                 ,  X  =  1  ±  V^^. 

EXERCISE  131 
Solve  the  following  equations : 

1.   (2a^-3a?)=^-8(2a^-3a;)  =  9. 


272  ALGEBRA 

'2.  a^H- 103^^4- 23  0.-2-100^-24  =  0. 

3.  x4-12ar^  +  14ar^+132aj-135  =  0.     %u 

•^4.  5  a;  +  12  +  5  V5  x  + 12  =  -  4. 

5    a^'-3        2  a;    ^     17 
*      2x        x'-S     '~  4* 

6.  V5^+I  +  3a/5¥+I  =  10. 

7.  3a;2  +  aj  +  5V3^T^T6  =  30. 

8.  8a.*2-l  +  6W8^^^1  =  -8a^. 

9.  a;*-2aaj3-17aV+18a3a;  +  72a^  =  0. 

/lO.  (7a;-6)^-5(7a;-6)*  =  -6," 

11     ^'  +  2      2d-5^35 
'  2  d  -  5      d^  +  2       6  ' 

^2.  £c2^7Va;2-4a)  +  ll  =  4a;-23. 

^.  Var^-3a;-3  =  a;2-3aj-23. 

14.  (2a;2-3a;-l)3-7(2a^-3i»-l)^  =  8. 

A5.  S-s/x'-12x-7-s/x'-12x  =  -2. 

^16,  7c'-lS](^  +  109k^-252k-\-lS0  =  0, 

17.  2ar'  +  4a;  +  Var^H-2x-3  =  9. 

18.  7  («3  _  28)-t  4-  8  (o^  -  28)-^  =  - 1. 

19.  (3^^  +  15)-* -5  (3  ^  +  15)-^  =  24. 

20.  9a;^-.12a;3_35^2^26a;  +  40  =  0. 

21    a:^-5a;  +  l      x^-2x-^2^     8 
*  ic2_2a;  +  2      aj2_5^_^-,^-      3* 

22.  9(x  +  a)^-22b\x-^a)^-\-Sb*  =  0, 

23.  ar^  +  l+Va;2-8ajH-37  =  8(a;4-12). 

24.  25  (a;  + 1)"^  - 15  (x  +  1)~^  =  -  2. 


25.  J?+?- JIZ 


THEORY  OF  QUADRATIC  EQUATIONS     273 


XXI.  THEORY  OF  QUADRATIC  EQUATIONS 

295.  Number  of  Roots.  JU„,<j^.  'T3u-tr-    Cl--*1  (jtvX,!^  A^'-rCv 
A  quadratic  equation  mrmot  have  more  than  two  different  roots. 
Every  quadratic  equation  can  be  reduced  to  the  form 

g,oif  +  bx-\-c  =  0.         ^-^v  ■»     '•*-« 

'^'^Tti  possible,  let  this  have  three  different  roots,  r^,  rg,  and  r^. 

Then,  by  §  81,  ar,'  +  br^  4-  c  =  0,  (1) 

ar2'  +  br2-{-c  =  0,  (2) 

and  ars'  +  brs-\-c  =  0,  (3) 

Subtracting  (2)  from  (1),  a  (?v  —  ^2^)  +  ^  (n  —  **2)  =  0. 

Then,  a  (r^  +  rg)  (rj  -  rg)  +  6  (rj  -  ^2)  =  0, 

or,  (ri  —  rg)  (an  +  a^a  +  6)  =  0. 

Then,  by  §126,  either  ri  — r2=0,  or  ari+ar2+6=0. 

But  ri  —  7'2  cannot  equal  0,  for,  by  hypothesis,  ri  and  rg  are 
•different. 

Whence,  ar^  +  arg  +  6  =  0.  (4) 

In  like  manner,  by  subtracting  (3)  from  (1),  we  have 

ar^  +  arg  +  6  =  0.  (5) 

Subtracting  (5)  from  (4),  ar2— a?3=0,  or  rg— r3=0. 

But  this  is  impossible,  for,  by  hypothesis,  ^v  and  r^  are 
different;  hence,  a  quadratic  equation  cannot  have  more  than 
two  different  roots. 

296.  Sum  of  Roots  and  Product  of  Roots. 

Let  ?'i  and  rg  denote  the  roots  of  ax^  +  6a;  +  c  =  0. 


By  §289,  ^^^-fe4:Vi-'-4ac^  ^^^  ^^^  _&_ V^ _4«c. 
2  CL  2i  a 


274  ALGEBRA 

Adding  these  values,  rj  +  rg  =  ^ —  = 

2a  a 

Multiplying  them  together,  ^ 

4a^  4a^     a 

Hence,  if  a  quadratic  equation  is  m  the  form  ax^  -{-hx-\-c  —  0, 
the  sum  of  the  roots  equals  minus  the  coefficient  of  x  divided  " 
by  the  coefficient  of  a^,  and  the  product  of  the  roots  equals  the 
independent  term  divided  by  the  coefficient  of  x^.   t 

1.  Find  by  inspection  the  sum  and  product  of  the  roots  of 

Jl       ** 

The  sum  of  the  roots  is  -,  and  their  product  ^^^ — ,  or  —  5.        9i^^^  3.    ft. 

2.  One  root  of  the  equation  6  a^  4- 31  a;  =  —  35  is  —  |-;  find 
the  other. 

The  equation  can  be  written  6  cc^  +  31  aj  4. 35  =  0. 

31 

Then,  the  sum  of  the  roots  is 

6 


31      /     7\  31      7 

Hence,  the  other  root  is (  —  ),  or 1--, 

6       \     2/'  6      2' 


or--. 


We  may  also  find  the  other  root  by  dividing  the  product  of  the  roots, 

^,by-L 
6      "^       2 

We  may  find  the  values  of  certain  other  expressions  which 
are  symmetrical  in  the  roots  of  the  quadratic. 

3.   If  ri  and  rg  are  the  roots  of  aa^  +  6a;  +  c  =  0,  find  the  value 
of  ri^  +  rjrg  +  ?'2^. 

We  have,  ri^  +  rir^  +  r2^  =  (n -\- rzY  -  rir2. 

b  c  * 

But,  n  +  ra  =  — ,  and  nr^  =  — 

a  a 

Whence,  rv'  +  r.r,  ^r,^  =  ^^-^  =  ^^^. 


THEORY  OF  QUADRATIC  EQUATIONS     275 

EXERCISE  132 

Find  by  inspection  the  sum  and  product  of  the  roots  of : 
>  1.    a;2  +  8aj4-7=0. -^.^      5.    2x-Ux'  =  7.   .'^    - 
>2.    x'  +  x-20  =  0.-l:-~     6.    10  +  12aj-15a^  =  0.  I 

3.  a.'2-6a;  +  l  =  0.  ^  >7.    Sx^-2  =  -x.  -^      ~  ^ 

4.  4a;2-.r-5  =  0.  4,  -  ^^'    8.    9 mV  +  21  w,?2|c)+ 5 n^  =  0. 
/'  9.    One  root  of  a;^  +  7  a;  =  98  is  7  ;  find  the  other.    S  -  -  - 

y\  10.    One  root  of  28  ar^  —  a;  —  15  =  0  is  —  f ;  find  the  other,  s  =  i^ 
(4,1.    One  root  of  5  a-^  -  17  a;  +  6  =  0  is  | ;  find  the  other,     ^^^"t 

If  Ti  and  7-2  are  the  roots  of  aoc^  -\- bx -{- c  =  0,  find  the  values 
of:       c^.^-^  a^ 
12.   ri+lL.       13.'i  +  i.       14.   1  +  -1.       15.   r,3  +  r/. 

297.   Formation  of  Quadratic  Equations. 

By  aid  of  the  principles  of  §  296,  a  quadratic  equation  may 
be  formed  which  shall  have  any  required  roots. 
For,  let  rj  and  r^  denote  the  roots  of  the  equation 

aaj2  +  6a;  +  c  =  0,  ora^^^^^c^O.  (1) 

a      a 

Then,  by  §  296,  -  =  -  rj  -  rg,  and  -  =  r^r^. 

Substituting  these  values  in  (1),  we  have 

sc^  —  riX  —  r^  +  TiTz  =  0. 

Or,  by  §  108,  (a;  -  r,)  (x  -  n)  =  0. 

Therefore,  to  form  a  quadratic  equation  which  shall  have 
any  required  roots. 

Subtract  each  of  the  roots  from  x,  and  place  the  product  of  the 
resulting  expressions  equal  to  zero. 

Ex,     Form  the  quadratic  whose  roots  shall  be  4  and  —  J. 


276  ALGEBRA 


By  the  rule,  {x-i){x+-]  =  0. 


Multiplying  by  4,  (x  -  4)  (4  a:  +  7)  =  0  ;  or,  4 x2  -  9x  -  28  =  0. 

EXERCISE  133 
Form  the  quadratic  equations  whose  roots  shall  be : 
"1.   5,8.  3.    -1,  -f.      5.   i,  -|.        -7.    -I,  -f. 

2.    -4,3.  4.   6, -L3.      ,6.   -1/,  0.  8.    -i,j\. 

9.    a-{-2b,  a-2b.  .11.    _4-|-5V3,  -4-5V3. 

VlO.   3m-n,m  +  An.        >12.    V^jh^V^,  V^-^V^K 

2  '  2 

V  ^     .    ^  .>  ^    .^  /  'j^     7  FACTORING 

298.   Factoring  of  Quadratic  Expressions. 
A  quadratic  expression  is  an  expression  of  the  form 
ax^  -\-hx-\-c. 

In  §  117,  we  showed  how  to  factor  certain  expressions  of  this 
form  by  inspection ;  we  will  now  derive  a  rule  for  factoring  any 
quadratic  expression ;  we  have, 

V  a      a) 

\_         a      \2aJ       4:a^     a  J 
W       2aJ  4.a'    J 

V  2a^        2a       J\   ^ 2a  2a       J' 

by  §  114. 
But  by  §  289,  the  roots  of  ax^  +  &a;  +  c  =  0  are 


^  _i.vV-4ac^„,        b       ^b^-4.ac 
2~a'^       2a        ^""^  ~21l        ^2^^ 


THEORY  OF   QUADRATIC   EQUATIONS  277 

Hence,  to  factor  a  quadratic  expression,  place  it  equal  to  zero, 
and  solve  the  equation  thus  formed. 

Then  the  required  factors  are  the  coefficient  of  x^  in  the  given 
expression,  x  minus  the  first  root,  and  x  minus  the  second. 

1.  Factor6a^  +  7i»-3. 

Solving  the  equation  6  a;2  +  7  a;  -  3  =  0,  by  §  289, 

-7i:Vi9T72^-7i:ll^l  ^^      3_ 
12  12  3  2 

Then,  .  6  a;2  +  7  a;  -  3  =  6 f  x  - 1  Vx  +  ^^ 

=  3(a:-|)x2^a;+|U(3:r-l)C2x  +  3). 

2.  Factor  4 4- 13  a?- 12 a^. 

Solving  the  equation  4  +  13  x  -  12  x^  =  0,  by  §  289, 


-  13  ±  V169  +  192  ^  -  13  ±  19  ^  _  1  ^j.  4 
-24  -24  4        3* 

Whence,  4  + 13a;  -  12a;2  =- 12(x +  -Vx --^ 

=K-i)x(-)e-l) 

=  (l  +  4«)(4-3a;). 

3.   Factor  2ar'-3a;2/-2/-7a;  +  42/  +  6. 
We  solve  2  a;2  -  a:(3y  +  7)- 2  ^2  +  4  ^  _!_  6  _  0. 


By  §  289, 


^3y  +  7±V(3y  +  7)2  +  16y2_32y-48 
4 


_  3  y  +  7  zb  V25  y2  +10  y  -|.  1  ^  3  y  +  7  ±  (5  y  +  1) 
4  4 

SJM^  or  =l^±i  =  2y  +  2or  ZliL±3. 
4  4^2 


Then,  2x2-3a;i/-2i/2_7ic  +  4?/  +  6 

=  2[x-(2y  +  2)][a;-~^  +  ^" 


2 

:(a;-2y-2)(2a;  +  y-3). 


^^278 

ALGEBRA 

/ 
Tac 

EXERCISE  134 

itor  the  following : 

1. 

0^4-14^  +  33. 

12. 

la^-lSax-j-Sa^. 

2. 

a^_13  0^4-40. 

13. 

21a^  +  5ic-6. 

3. 

a^-a-42. 

14. 

12ic2_l6a;-35. 

4. 

2x'-^llx-6. 

15. 

3  4.13a;_30a;2. 

5. 

4a^  +  19a;  +  12. 

16. 

28-5^-12^1 

6. 

8m2-14m  +  5. 

17. 

16x^'-\-S0x-\-9, 

7. 

Q-Sx-af, 

18. 

18a^-31aj  +  6. 

8. 

10a^  +  39x  +  14. 

19. 

6a^-23a;-35. 

9. 

40  4-19a;-3a^. 

20. 

20-13  a;-15aj2. 

10. 

SB-x-Qx". 

21. 

20a^-^SSmx-\-7m\ 

11. 

5a^-21nx-\-lSnK               22. 

24:x'-3Sxyz-\-15yh'. 

23.  x^-xy-6y^-6x-}-13y-\-5. 

24.  i»2_3a;y_42/2  4_e^_4^4_3^ 

^       25.  x^-6xy  +  5y'-2x-2y-S. 

26.  2a2  +  5a6  +  262  +  7a.4-5&  +  3. 

27.  3x^-\-Txy-6y^-10x-Sy-{-S. 

28.  2-72/-7a;  +  3^2  4.^2/-4a;2. 

29.  6a;2-2a;2/-202/'  +  5a;2  +  232/2;-622. 

299.  If  the  coefficient  of  x^  is  a  perfect  square,  it  is  prefer- 
able to  factor  the  expression  by  completing  the  square  as  in 
§  287,  and  then  using  §  114. 

1.   Factor  9a^-9a;-4. 

By  §  287,  9  a;2  —  9  a;  will  become  a  perfect  square  by  adding  to  it  the 

9  ^ 

square  of  ,  or  — 

2x8'       2 


THEORY   OF   QUADRATIC   EQUATIONS  279 

Then,  q x^  -9 x- ^  =  9 x"^ -9 x  +  (^Y ---4:=  ^8^:--^-- 


(^^-M)(«^-M)     «^^*> 


=  (3x  +  l)(3x-4). 

If  the  oc^  term  is  negative,  the  entire  expression  should  be 
enclosed  in  parentheses  preceded  by  a  —  sign. 

2.   Factor  3  -  12  a;  -  4  a^. 

3  _  12  X  -  4  a:2  =  -(4  a:2  +  12  a;  -  3) 

=  -  (4  a:2  +  12  X  +  9  -  9  -  3) 

=  -[(2x  +  3)2_12] 

=  (2  X  +  3 -f- VT2)  X  (- 1)(2  X  +  3  -  VT2) 

=  (2V3  +  3-f2x)(2V3-3-2x). 

EXERCISE  135 
Factor  the  following : 

1.  ^x'-Ux-l.       .  7.   l-f-2a;-a^. 

2.  9a^-21a;  +  10.  8.    16  x^  -  16  x -\- 1. 

3.  ar^  +  a;-12.  9.    6~5x-25x'. 

4.  16a^  +  40a;  +  21.  10.   4.x^  +  9x-9. 

5.  9.^2  +  2405-2.  11.   36  a:^  _|_  72  a^ -I- 29. 

6.  4x2  +  20a;  +  19.  12.   25a^-10x-ll. 

300.  We  will  now  take  up  the  factoring  of  expressions  of 
the  forms  x'^  +  aa^y^  -f-  y^,  or  x*^  -\-  y*,  when  the  factors  involve 
surds.     (Compare  §  115.) 

1.   Factor  a^  +  2  a'b^  +  25  b\ 

a*  +  2  a262  +  25  64  =  (a*  +  10  a262  +  25  &4)  -  8  a262 
=  (a2+5  62)2_(a6V8)2 
=  (a2  +  5  &2  +  a&V8)(a2  +  5  ^2  _  ahVS) 
=  (a2  +  2  a&\/2  +  5  62)  (^2  _  2  ahV2  +  5  62).  * 


280  ALGEBRA 

2.  Factor  x*  +  l. 

=  (X2+1)2-(XV^)2 

=  (a;2  +  X  V2  +  1)  (a;2  _  a,  V2  + 1). 

EXERCISE  136 

In  each  of  the  following  obtain  two  sets  of  factors,  when 
this  can  be  done  without  bringing  in  imaginary  numbers : 

1.  x^-7a^-h4:.  4.   4a4  4-6a2  +  9. 

2.  a'-{-b\  5.   36«^-92a^  +  49. 

3.  Qm^-llm^  +  l.  6.   25 m^  +  28 mV  + 16 w^ 

301.   Solution  of  Equations  by  Factoring. 

In  §  126,  we  showed  how  to  solve  equations  whose  first  mem- 
bers could  be  resolved  by  inspection  into  first  degree  factors, 
and  whose  second  members  were  zero. 

We  will  now  take  up  equations  whose  first  members  can  be 
resolved  into  factors  partly  of  the  first  and  partly  of  the 
second,  or  entirely  of  the  second  degree. 

1.    Solve  the  equation  x^  +  1  =  0. 

Factoring  the  first  member,     (x -\- 1)  (x^  —  x -j- 1)  =  0. 

Then,  ic+l=0,  ora;  =  -l; 

and    x2  - x  +  1  =  0  ;  whence,  by  §  289,  x  =  1  ±  ^1  - ^  =  l±y/-S^ 

2:   Solve  the  equation  a;*  + 1  =  0. 

.By  Ex.  2,  §  300,  x^  +  1  =  (a;2  +  x\/2  +  1)  (x2  -  x\/2  +  1). 

Solving  x2  +  x\/2  +  1  =  0,  we  have 

^  ^  -V^  J:V2"^^  _  -  V2  J:  \/^r2_ 
2  2 

Solving  x2  -  x\/2  +  1  =  0,  we  have 

^  ^  V2  ±y/2^^^  \/2  -J:  V^ 

2  2 


THEORY   OF   QUADRATIC   EQUATIONS  281 

The  above  examples  illustrate  the  important  principle  that  the  degree 
of  an  equation  indicates  the  number  of  its  roots  ;  thus,  an  equation  of  the 
third  degree  has  three  roots  ;  of  the  fourth  degree,  four  roots  ;  etc. 

The  roots  are  not  necessarily  unequal ;  thus,  the  equation  x^— 2  x+l  =  0 
may  be  written  (x  —  l){x  —  \)  =  0,  and  its  two  roots  are  1  and  1. 

EXERCISE  137 

Solve  the  following  equations : 

1.   5a;3-3a;2-9a;  =  0.  11.  a^  -  6a.-2  + 1  =  0. 

/2.    (aj-h4)(2x2_^5a?  +  25)=0.  12.  x^-ox'  +  l^O, 

V3.    (9x2-4)(llar^+8a;-4)=0.  13.  e>4.:^ -125  =  0. 

4.  a;4-llaV-12a*  =  0.  14.  x^- 10^2  +  9  =  0. 

5.  a;^-81  =  0.  15.  a;^- 20x^4- 16  =  0. 

6.  a;^  +  2a^  +  2aj  +  4  =  0.  16.  9a;*  +  5a;2  +  4  =  0. 
1/7.    x«-l  =  0.  17.  a;6-729  =  0. 

1^.   x^  +  8aj  =  0.  18.  0^8-256  =  0. 

9.   5a)3-4a.'^  +  60a.-48  =  0.  a,^4-2a^  +  4^4 

10.   2l7?  +  ^a^  =  0.  '  a^-2a;-4     x" 
20.   9aj*-a;2-f  4  =  0. 


21.    ■Va;^  +  l4-V9a;4-a^  =  2a;-l. 

302.   Discussion  of  General  Equation. 

By  §  289,  the  roots  oi  ax^  -\- hx -\- c  =  0  are 


2a  2a 

We  will  now  discuss  these  results  for  all  possible  real  values 
of  a,  b,  and  c. 

I.  6^  _  4  (5^c  positive. 

In  this  case,  Vi  and  rg  are  real  and  unequal. 

II.  62_4ac  =  0. 

In  this  case,  rj  and  rg  are  ?'ea/  and  egwaZ. 


III.  6^  —  4  ac  negative. 

In  this  case,  rj  and  rg  are  imaginary  (§  273).     £^-^1.    -n— ^rw^JL 

IV.  6  =  0. 

In  this  case,  the  equation  takes  the  form  ,- — 

ax^  +  c  =  0  J   whence,  a;  =  ±  -y '  ' ' 

If  a  and  c  are  of  unlike  sign,  the  roots  are  real,  equal  in  abso- 
lute value,  and  unlike  in  sign. 

If  a  and  c  are  of  like  sign,  both  roots  are  imagi7iary. 

V.  c  =  0. 

In  this  case,  the  equation  takes  the  form  ^  *  ~  ^^  U-*V 

aa^-|-6a;  =  0;   whence,  a;  =  0  or .^ '^ 

a 

Hence,  the  roots  are  both  real,  one  being  zero.  *^ 

VI.  6  =  0,  and  c  =  0. 

In  this  case,  the  equation  takes  the  form  ax^  =  0.      '^  r    ^  ^y' 
Hence,  both  roots  equal  zero.  ^^ 

The  roots  are  both  rational,  or  both  irrational,  according,  as 
6^  —  4  ac  is,  or  is  not,  a  perfect  square. 

Ex.     Determine  by  inspection  the  nature  of  the  roots  of 

2x2_5ir-18  =  0. 

Here  a  =2,  &  =  -  5,  c  =  -  18  ;   and  62  _  4  ^c  =  25  +  144  =  169. 
Since  h^  —  i^ac  is  positive,  the  roots  are  real  and  unequal. 
Since  h^  —  4iac  is  a  perfect  square,  both  roots  are  rational. 

EXERCISE   138 

Determine  by  inspection  the  nature   of  the   roots   of  the 
following:  ,,JL,>^,  _,    ,v 

1.  &x'  +  llx-\-^  =  0.  /61  6.   aj2_i9a;4. 125  =  0.  -if 

2.  6aj2  +  a;  =  o:"..^....  ;.-^.--^    7.   ^x'-4.x  =  0.     i^^'tS 
0^  ~  ^(>  AM>3r 


4.  12«2-19«  +  4  =  0.''^:^    ^^J^-   16a;2  4.24a;H-9  =  0.    ^ 

5.  25a;2_4=:0.  >^. 11.^0.    30a^-30  =  llaj.     J2  7  2. 


THEORY  OF   QUADRATIC   EQUATIONS 


283 


GRAPHICAL    REPRESENTATION    OF    QUADRATIC    EXPRES- 
SIONS WITH  ONE  UNKNOWN  NUMBER 

303.  The  graph  of  a  quadratic  expression,  with  one  unknown 
number,  x,  may  be  found  by  putting  y  equal  to  the  expression, 
and  finding  the  graph  of  the  resulting  equation  as  in  §  181. 

1.   Find  the  graph  of  x^  —  2  a;  —  3. 
Put  y  =  ic2  -  2  a;  -  3. 


If  X  =  0, 

2/ =  -3.     {A) 

If  x=l, 

y  =  -4.     (B) 

If  X  =  2, 

y  =  -s.    (C) 

lix  =  3, 

y  =  o.       (D) 

If  a;  =  4, 

y  =  5.       (E) 

If  a:  =  -  1, 

y  =  0.       (F) 

If  a;  =  -2, 

y  =  5.      (G) 

The  graph  is  the  curve  GBE. 

By  taking  other  values  for  a;,  the  curve  may  be  traced  beyond  E  and  0 ; 
it  extends  in  eitlier  direction  to  an  indefinitely  great  distance  from  XX'. 

[We  may  determine  the  lowest  point  of  the  curve  hy  the  artifice  of  com- 
pleting the  square,  as  follows  : 

We  have,  x^  -  2  x  -  3  :=  (a:^  -  2  a;  +  1)  -  1  -  3  =  (x  -  1)2  -  4. 

The  latter  expression  has  its  negative  value  of  greatest  absolute  value 
when  X  =  1,  being  then  equal  to  —  4. 

Then,  the  lowest  point  of  the  curve  has  the  co-ordinates  (1,  —  4);  and 
is  therefore  the  point  B']. 

2.   Find  the  graph  of  2x2  + a; -3. 
Put  y  =  2x2  +  x-3. 

If  X  =  0,       y=-S.' 

If  X  =  1,       y  =  0.. 

If  X  =  2,       y  =  7. 

If  x  =  -l,  y=-2. 

If  X  =  -  2,  y  =  3. 
The  graph  is  the  curve  ABC. 
[To  find  the  lowest  point  of  the  curve,  we  have 

2x2  +  x-3  =  2fx2  +  ^V3  =  2fx2  +  ^+i-^-A_3  =  2fx  +  -y-— . 
V         2^  V         2      16;      16  V        4/        8 


284 


ALGEBRA 


The  latter  expression  has  its  negative  value  of  greatest  absolute  value 


1  25 

when  X  =  — ,  being  then  equal  to 

4  8 

Then,  the  lowest  point  has  the  co-ordinates 


)•] 


304.  The  principle  of  §  188  holds  for  the  graph  of  the  first 
member  of  any  quadratic  equation,  with  one  unknown  number. 

Thus,  the  graph  of  a^  —  2  x  —  3  (§  303)  intersects  the  axis 
XX'  at  points  whose  abscissas  are  3  and  —  1,  and  the  equation 
ic^  —  2a7  —  3  =  0  has  the  roots  3  and  —  1. 

Again,  the  graph  of  2  x^-{-x  —  3  intersects  XX'  at  the  point 
whose  abscissa  is  1,  and  between  the  points  whose  abscissas  are 
—  1  and  —  2 ;  and  the  equation  2  cc^  +  a?  —  3  =  0  has  one  root 
equal  to  1,  and  one  between  —  1  and  —  2. 

EXERCISE  139 

Find  the  graph  of  the  first  member  of  each  of  the  following 
equations,  and  verify  the  principle  of  §  188  in  the  results : 

1.  x^-5x-\-A  =  0.  5.  4.x'-^Tx  =  0. 

2.  x^-\-x-6  =  0.  6.  2a^-lla;-6  =  0. 

3.  a;2-f-7a;  +  10  =  0.  7.  6x2  +  5a;-6  =  0. 

4.  Sx'-4:X  =  0.  8.  Sx'-Ux-15  =  0. 


305.   Graphs  of  the  First  Members  of  Quadratic  Equations  hav- 
ing Equal  or  Imaginary  Roots. 

1.    Consider  the  equation  a?^  —  4  a?  +  4  =  0. 

We  may  write  the  equation  (a;  — 2)(x— 2)  =  0. 
Then,  by  §  126,  the  roots  are  2  and  2. 
To  find  the  graph  of  the  first  member,  put 
y  =  (x-  2)2. 

If    x  =  0,  y  =  4.  If    x  =  2,  y  =  0. 

If    x  =  l,  y  =  l.  If    x  =  S,  y  =  l;  etc.     X- 

The  graph  is  the  curve  ABC,  which  extends 
to  an  indefinitely  great  distance  from  XX'. 

Since  (x  —  2)2  cannot  be  negative  for  any  value  of  x,  y  cannot  be  nega- 
tive ;  and  the  grapli  is  tamjent  to  XX'. 


THEORY  OF  QUADRATIC  EQUATIONS 


285 


It  is  evident  from  this  that,  if  a  quadratic  equation,  with  one 
unknown  nuinber,  has  equal  roots,  the  graph  of  its  first  member 
is  tangent  to  XX'. 

2.   Consider  the  equation  ic^  +  a;  +  2=0. 


Solving, 


- 1  j-  V-  7 


!>  i 

•^  o 

r 


To  find  the  graph  of  the  first  member,  put 

1/  =  x2  +  X  +  2. 

If  x  =  0,2/ =  2.        Ifa;=-1,  y  =  2. 

If  X  =  1,  y  =  4.        If  X  =  -  2,  y  =  4  ;  etc. 

The  graph  is  the  curve  ABC,  which  extends 
to  an  indefinitely  great  distance  from  XX'. 

We  have,     x2  +  x  +  2  =  (x^  +  x  +  ^^  -  ^  +  2  =  (x  +  ^V  +  ^• 

Since    (x  +  -  )   +-  cannot  be  zero  or  negative  for  any  value  of  x,  y 
V-  ,  2/       4 

cannot  be  zero  or  negative,  and  the  graph  does  not  intersect  XX'. 

It  is  evident  from  this  that,  if  a  quadratic  equation,  with  one 
unknown  number,  has  imaginary  roots,  the  graph  of  its  first 
member  does  not  intersect  XX'. 


EXERCISE  140 

Find  the  graphs  of  the  first  members  of  the  following,  and 
in  each  case  verify  the  above  principles : 

2.   a^  +  3a;-f-4  =  0.  4.   2x^-Ax  +  5  =  0. 


286  ALGEBRA 


XXII.    SIMULTANEOUS    QUADRATIC 
EQUATIONS 

306.  On  the  use  of  the  double  signs  ±  and  T . 

If  two  or  more  equations  involve  double  signs,  it  will  be 
understood  that  the  equations  can  be  read  in  two  ways ;  tirstj 
reading  all  the  n2:>per  signs  together;  second,  reading  all  the 
lower  signs  together. 

Thus,   the   equations  x  =  ±2,  y  =  ±3,  can  be  read   either 

x  =  -^2,y  =  +  3,OTx  =  -2,y  =  -3. 
Also,    the   equations   x  =  ±2,  y=:f  3,  can  be   read   either 
x  =  -j-2,y  =  -3,ovx  =  -2,y  =  -{-3. 

307.  Two  equations  of  the  second  degree  (§  83)  with  two 
unknown  numbers  will  generally  produce,  by  elimination,  an 
equation  of  the  fourth  degree  with  one  unknown  number. 

Consider,  for  example,  the  equations 

fx^^y=a.  (1) 

\x-\-f  =  b.  (2) 

From  (1),  y  =  a  —  x^',  substituting  in  (2), 
x-\-a^  —  2  a'y? -\-x^—h\ 

an  equation  of  the  fourth  degree  in  x. 

The  methods  already  given  are,  therefore,  not  sufficient  for 
the  solution  of  every  system  of  simultaneous  quadratic  equa- 
tions, with  two  unknown  numbers. 

In  certain  cases,  however,  the  solution  may  be  effected. 

308.  Case  I.     When  each  equation  is  in  the  form  ■..   , 

ax^  +  by-  =  c. 

In  this  case,  either  cc-  or  y^  can  be  eliminated  by  addition  or 
subtraction. 


SIMULTANEOUS   QUADRATIC   EQUATIONS        287 


1.    Solve  the  equations  i„   ^     ^^    ^ 

1 3  ?/2  — 11  03^  =  4. 

(1) 

(2) 

Multiply  (1)  by  3,                      9x^  +  12  y^  =  228. 
Multiply  (2)  by  4,                      12  y^-  44  x2  =    16. 

Subtracting,                                              53  x^  =  212. 

Then,                                                           x^  =  4,  and  a;  =  ±  2. 

Substituting  x  =  ±  2  in  (1),            12  +  iy^  =  76,  or  ^y'^  =  64. 

Then,                                                           y^  =  16,  and  2/  =  ±  4. 

The  solution  is  x  =  2,  ?/  =  ±  4  ;  or,  x  =  —  2,  ?/  =  ±  4. 

In  this  case  there  are  four  possible  sets  of  values  of  x  and  y  which 
satisfy  the  given  equations  : 

1.  x  =  2,  ?/  =  4.  3.   x  =  -2,  y  =  4. 

2.  x  =  2,  y=-4.  4.    x  =  -2,  ?/  =  -4. 

It  would  not  be  correct  to  leave  the  result  in  the  form  x  =  ±  2,  y  =  ±  4, 
for  this  represents  only  the  first  and  fourth  of  the  above  sets  of  values. 

The  method  of  elimination  by  addition  or  subtraction  may  be 
used  in  other  examples. 

.«     ^  ,        ,  .         r3a^-42/  =  47.  (1) 

"  2.    Solve  the  equations  i       „     ^  ^  ,/^ 

^  l7aj2  +  62/  =  33.  (2) 

Multiply  (1)  by  3,  9  x2  -  12  y  =  141. 

Multiply  (2)  by  2,  Ux^+12y=   66. 

Adding,  23x2  =  207. 

Then,  x2  =  9,  and  x  =  ±  3. 

Substituting  x  =  ±  3  in  (1),  27  —  4  ?/  =  47,  and  ?/  =  —  5. 

It  is  possible  to  eliminate  one  unknown  number,  in  the  above  examples, 
by  substitution  (§  169),  or  by  comparison  (§  170). 


EXERCISE  141    - 

/^  ISolve  the  following  equations : 

(Sx'-\-2y'  =  66.  (Sx-5y'  =  -116. 

l9a^  +  52/'  =  189.  "      '     l7a;  +  4/  =  121. 


288  ALGEBRA 


fll(c2-62/'  =  84. 
4.    -^  ^  _     ^_, „     „  7. 


3  a;2  -  2  iK.V  =  24. 
3.     1  6. 

4:X^  —  5xy  =  46. 


8/i2-27A:2=:6.  I     .^2  _^  a:?/ +  3  2/' =  27. 

la^_32/2  =  _95.  •     1  a^-2/  =  17. 

4  a;2  _  ^2  ^  3  ^2  ,  10  a6  +  3  62. 


5. 


9.     , 

42/'-a;2  =  3a--10a64-362. 

309.  Case.  II.  When  one  equation  is  of  the  second  degree, 
and  the  other  of  the  first. 

Equations  of  this  kind  may  be  solved  by  finding  one  of  the 
unknown  numbers  in  terras  of  the  other  from  the  first  degree 
equation,  and  substituting  this  value  in  the  other  equation. 

2  cc^  —  xy  =  6y.  (1) 

x+2y  =  7^  (2) 

From  (2),  2y  =  7  -  x,  or  y  =  '^-^^ -  (3) 


Ex,    Solve  the  equations 


Substituting  in  (1),  2x^-x( - — - \  =  6 [ 'L^^ ]  • 

Clearing  of  fractions,  4  a;2  -  7  a;  +  x^  =  42  -  6  ib,  ot  6x^-x=i2. 

Solving,  x  =  S  or  —  — . 

Substituting   in    (3),    y  =  ^-?  or  ——2.  =  2  or  — . 
2  2  10 

The  solution  isa;  =  3,  w  =  2:  or,  a;  = ,w  =  — 

Certain  examples  where  one  equation  is  of  the  third  degree  and  the 
other  of/the  first  may  be  solved  by  the  method  of  Case  II. 

y  ■  ' 

y  EXERCISE  142 

1»Solve  the  following  equations : 
/^       ^      ra^4-32/2  =  37.  ^      p  +  2/  =  _4. 

la;— 2?/ =9.  *     1      ic?/  =  — 45. 


SIMULTANEOUS  QUADRATIC   EQUATIONS         289 


3. 


5. 


X—  y  =  5. 
x^-xy-\-2y^  =  8. 
3x-\-y  =  10, 

X  -\-  y  =  7. 
x^  —  xy-\-y^  =  124. 
a;  +  2/=8. 

,     2e-3^  =  5. 
344. 

y 

x  —  y  =  a-\-2h. 
[o^  +  y''  =  o?  +  2ab-\-2h\ 


rar^- 2/3  =  34 
U  -w  =8. 


10. 


f2a;      3?/_o 


12a;     3^ 


2. 


a;?/  =  a^  +  a  —  2. 

3x  +  42/  =  7a  +  2. 

a;         40 
21 


310.  Case  III.  When  the  given  equations  are  symmetrical 
with  respect  to  x  and  y ;  that  is,  when  x  and  y  can  be  interchanged 
without  changhig  the  equation. 

Equations  of  this  kind  may  be  solved  by  combining  them  in 
such  a  way  as  to  obtain  the  values  oi  x-\-y  and  x  —  y. 


1.   Solve  the  equations      \ 


x-\-y=:2. 
xy=^  — 15. 


Squaring  (1),                    x^ 

+  2  5cy  +  ?/2  =        4. 

Multiplying  (2)  by  4, 

4  xy          =  -  60. 

Subtracting,                      x2 

-  2  xy  +  y2  =      64. 

Extracting  square  roots, 

x-2/=±8. 

Adding  (1)  and  (3), 

2  a;  =  2  ±  8  =  10  or  -  6. 

Whence, 

X  =  6  or  -  3. 

Subtracting  (3)  from  (1), 

2  y  =  2  =F  8  =  -  6  or  10. 

Whence, 

?/  =  -  3  or  5. 

The  solution  is  a;  =  5,  y  = 

-  3  ;  or,  a;  =  -  3,  y  =  h. 

(1) 

(2) 


(3) 


290  ALGEBRA       . 

In  subtracting  ±  8  from  2,  we  have  2  T  8,  in  accordance  with  the  nota- 
tion explained  in  §  306. 

In  operating  with  double  signs,  ±  is  changed  to  =F ,  and  T  to  ± ,  when- 
ever 4-  should  be  changed  to  — . 

(The  above  equations  may  also  be  solved  by  the  method  of  Case  II ;  but 
the  symmetrical  method  is  shorter  and  neater.) 

,         .  .  (x'  +  i/  =  50.  (1) 

2.  Solve  the  equations     i 

^  I         xy  =  -7.  (2) 

Multiply  (2)  by  2,  2xy=-  14.  (3) 

Add  (1)  and  (3),  x'^ +  2xy  +  y^  =  S6,  or  x+ y  =±6.  (4) 

Subtract  (3)  from  (1),  x^ -2xy  +  y^  =  64,  or  x  -  y  =±8.  (5) 

Add  (4)  and  (5),  2  x  =  6  ±  8,  or  -  6  ±  8. 

Whence,  a;  =  7,  —  1,  1,  or  —  7. 

Subtract  (5)  from  (4),  2  ?/  =  6  T  8,  or  -  6  T  8. 

Whence,  y  =  —l,  7,   —7,  or  1. 

The  solution  is  a?  =  ±  7,  2/  =  =F  1  ;  or,  a;  =  ±  1,  y  =^7. 

Certain  examples  in  which  one  equation  is  of  the  tJiird  degree, 
and  the  other  of  the  first  or  second,  may  be  solved  by  the 
method  of  Case  III. 

(a^-f  =56.  (1) 

3.  Solve  the  equations     i    .  „ 

^  \^  +  xy-\-f  =  2S.  (2) 

Divide  (1)  by  (2),  x-y  =  2.  (3) 

Squaring  (3) ,  x'^  -  2  xy  -\- y^  =  4.  (4) 

Subtract  (4)  from  (2),  Sxy  =  24,  or  xy  =  8.  (5) 

Add  (2)  and  (5),  x^  +  2  xy  +  y^  =  S6,  or  x  +  y  =±  6.  (6) 

Add  (3)  and  (6),  2ic  =  2±6  =  8  or  -4. 

Whence,  a;  =  4  or  —  2. 

Subtract  (3)  from  (6),  2  y  =  ±  6  -  2  =  4  or  -  8. 

Whence,  y  =  2  or  —  4. 

The  solution  is        x  =  i,  y  =  2  ;  or,  x  =  —  2,  y  =  —  i.   ' 
(If  we  interchange  x  and  y  in  equation  (1),  it  becomes 
y'^  —  x^  =  56,  or  x^  —  y^  =     56, 

which  is  not  the  same  as  (1). 


SIMULTANEOUS  QUADRATIC   EQUATIONS        291 


Thus,  the  equation  (1)  is  not  symmetrical  with  respect  to  x  and  y  ;  but 
the  method  of  Case  111  may  often  be  used  when  either  or  both  of  the  given 
equations  are  symmetrical,  except  with  respect  to  the  signs  of  the  terms.) 

We  may  advantageously  use  the  method  of  Case  III  to  solve 
certain  equations  which  are  not  symmetrical  with  respect  to  x 
and  2/;  as,  for  example,  the  equations 

ic  -  2  ?/  =  -  4. 
iB2  +  42/'  =  40. 


EXERCISE  143 

Solve  the  following  equations  by  the  symmetrical  method 
rx2  +  2/2  =  29. 


2. 


'I 


5. 


X  +2/  =-3. 
05  — 2/ =  11. 

a^  =  -28. 

g2+s2^130. 
q  -s  =-8. 

xy  =  12. 
x^-y^  =  35. 
,a?-\-xy^y'^  =  l. 


^      ar^-2/3  =  26. 
U  -  2/  =  2. 


I       xy  = 


8. 


2n  -1. 

xy  =  n"  —  n  —  2. 
x^ -\- xy  -\- y^  =  63. 
x-~y  =  S, 


x" 


10. 


11. 


•m: 


16. 


xy  +  f  =  a^-{-3b\ 
x-\-y  =  2a. 
a^  +  2/3^280. 
[x^-xy-^y^  =  2S. 
x^  +  xy-\-y^  =  l. 
^  x^  —  xy -\- y^  =  19.   ■ 
^3  + 71^  =  407. 

+  71=11. 
x'  +  9y'  =  50. 
x-3y  =  0. 
xy  =  —  16. 
I  2  a;  +  2/  =  14. 
36^2  +  64  2/' =  85. 
ex  +   8  2/  =  11. 
x^-Sy^  =  lS9. 
X  -2y  =9. 


311.  Case  IV.  When  each  equation  is  of  the  second  degree, 
and  homogeneous;  that  is,  ichen  each  term  involving  the  unknown 
numbers  is  of  the  second  degree  with  respect  to  them  (§  65). 


292  ALGEBRA 

Certain  equations  of  tliis  form  may  be  solved  by  the  method  of  Case  I 
or  Case  III.     (See  Exs.  1,  §  308,  and  2,  §  310.) 

The  method  of  Case  IV  should  be  used  only  when  the  example  cannot 
be  solved  by  Cases  I  or  III. 

x'-2xy  =5.  (1) 

0^  +  ^  =  29.  (2) 

Patting  in  the  given  equations  y  =  vx, 


Ex.     Solve  the  equations      \ 


we  have  x^  —  2vx^=    6:  or,  x^  = •  •  (3) 

9Q 

and  «2  4.  ^2x2  =  29  ;  or,  x"^ 


K  oq 

Equating  values  of  x^,  — - —  =  ,  or  5  v^ -{■  5S  v  =  2i. 

1  —  2  ■«     1  +  ^2 

2 

Solving  this  equation,  ?j  =  -  or  —  12. 

5 

Substituting  these  values  in  (3),  we  have 

^      or -— ^  =  25or  -;  then,  x=±5  or  ±    -^ 


1_4        1  +  24  5'  -  ^V6 

5 


Substituting  the  values  of  v  and  x  in  the  equation  y  =  vx, 

2.  ,  ^N  _      -,0/  ,     1   \_  ,  o..-^i2_ 

V6 


y  =  |(±5)  or  -12/±— W±2orT 


The  solution  isx  =  ±6,  y  =±2;  or,  a;  =  ± -i^,  y  =  T  — • 

In  finding  y  from  the  equation  y  =  vx,  care  must  be  taken  to  multiply 
each  value  of  x  by  the  value  of  v  which  was  used  to  obtain  it. 

The  given  equations  may  also  be  solved  as  follows  : 

Dividing  (1)  by  (2),  ^^~^^J^  =  — ,  or  29  a;2  -  58  a;w  =  5  x2  +  5  y2. 
3.2  4. 2/2       29 

Then,         5y2  +  58a;y  -  24a;2  =  0,  or  (5  2/ -  2x)(2/ +  12  x)  =  0. 

Placing  5y-2x=:0,  y  =  —  ;  substituting  in  (1), 

5 

a;2  _  1^  =  5^  or  x2  =  25. 
5 

Then,  x  =  ±  5,  and  y  =  ?-5=  ±2. 

5  ' 


SIMULTANEOUS   QUADRATIC   EQUATIONS         293 

Placing  2/  +  12  tc  =  0,  y  =  -V2x;  substituting  in  (1), 

ic2  4-  24  x2  =  5,  or  x^  =  -• 
5 

Then,  x  =  ±-!-,  and  ?/ =  -12a:  =  T--- 

V6  V5  ^ 

EXERCISE  144 

Solve  the  following  equations: 
I  aj2  +  ^2  =  25,  rp2^pg-5g2  =  25. 

fa;2  +  3a;?/  =  -5.  f  i«2_  2  a;?/ -4  2/^  =  - 41. 

l2a^2/-2/'  =  -24.  '    [x'-5xy -{-Sy'  =  5S. 
r  5x2-2/2  =  9.  |2x2  +  7x2/  +  42/2  =  2. 

lx?/-3/  =  -90.  *   l3x2  +  8x?/-4/  =  -72. 
ra;2^^^^2/2  =  19.  r4x2-2x2/-2/'  =  -16. 

I    2a^  +  x2/    =-2.  *  1  5x'-7xy  =  -36. 

^'   I         Sxy  +  y'  =  2S.  '   iBa^-Sxy -72y'  =  3S. 

312.    Special  Methods  for  the  Solution  of  Simultaneous  Equa- 
tions of  Higher  Degree. 

No  general  rules  can  be  given  for  examples  which  do  not 

come  under  the  cases  just  considered;    various  artifices  are 

employed,   familiarity   with   which    can    only  be    gained    by 

experience. 

[      :x^-f  =  19.  (1) 

1.    Solve  the  equations    i    „  <,     ^  /on 

[xy  —  xy^  =  6.  {Z) , 

Multiply  (2)  by  3,  Zx'^y-S  xy^  =  18.  (3) 

Subtract  (3)  from  (1),  x^-Sx'^y  +  3  xy'^  -y^  =  l. 

Extracting  cube  roots,  x  —  y  =  1.  (4)^ 

Dividing  (2)  by  (4),  xy  =  6.  (5) 

Solving  equations  (4)  and  (5)  by  the  method  of  §  310,  we  find  x  =  S, 
y  =  2;  or,  X  =  -  2,  y  =  -  3. 


294  ALGEBRA 

r*     ^  1        ,  .  ( x^-{-y^  =  9xy. 

2.  Solve  the  equations    i 

[    x-\-y  =  6. 

Putting  x  =  u  +  V  and  y  =  u  —  v, 
(u  +  vy  +  (w  -  vy  =  9(m  +  V)  (u  -  v),  or  2  m8  +  6  uv"^  =  9(^2  -  ^2)  ;  (1) 
and  '         (u -\- v)-\-{u —  v)=6,2u  =  6,  or  u  =  3. 

Putting  w  =  3  in  (1),     54  +  18  ^2  _  g^g  _  ^2). 

Whence,  v"^  =  1,  or  -y  =  ±  1. 

Therefore,  x  =  u  +  v  =  S±l  =  4:or2; 

and  yz=u  —  v  =  ST'i-=2  or  4. 

The  solution  is  x  =  i,  y  =  2  ;   or,  x  =  2,  y  =  i. 

The  artifice  of  substituting  ii  +  v  and  u  —  v  for  x  and  y  is  advantageous 
in  any  case  where  the  given  equations  are  symmetrical  (§  310)  with 
respect  to  x  and  y.     See  also  Ex.  4. 

3.  Solve  the  equations    -I  '  ^ 

I  xy  =  i^.  (2) 

Multiplying  (2)  by  2,  2xy  =  \2.  (3) 

Add  (1)  and  (3),      a;2  +  2  a;?/  +  ?/2  +  2  a;  +  2  ?/  =  35. 

Or,  (a;  +  y)2  +  2(a:  +  y)=35. 

Completing  the  square,  (x  +  ?/)2  +  2(ic  +  y)+ 1  =36. 

Then,  (x  +  y)  +  1  =  ±  6  ;  and  a;  +  y  =  5  or  -  7.  (4) 

Squaring  (4) ,  x2  +  2  xy  +  ?/2  =  25  or  49. 

Multiplying  (2)  by  4,  4a;y         =24. 


Subtracting,  x^-2xy  +  y'^  =  \  or  25. 

Whence,  a;  -  y  =  ±  1  or  ±  5.  (5) 

Adding  (4)  and  (5),  2  x  =  5  ±  1,  or  -  7  i  5. 

Whence,  x  =  3,  2,   -1,  or  -6. 

Subtracting  (5)  from  (4),  2  ?/  =  5  T  1,  or  -  7  T  5. 

Whence,  2/ =  2,  3,  -6,  or  -1. 

The  solution  ls.x=3,  y=2;  x=2,  ?/=3  ;  x=-l,  2/=-6;  or  x=-6, 
?/  =  -!. 

/I      a  1       XT,  .•  fiC^  +  2/^  =  97. 

4.    Solve  the  equations  \ 

Ix  -\-y  =  —  1. 


SIMULTANEOUS  QUADRATIC   EQUATIONS        295 


Putting  x=u  +  v  and  y  =  u-v, 

(u  +  vy  -\-(u-  vy  =  97,  or  2  m*  +  12  mV  +  2  v*  =  97, 


and 


Xu  +  v)-\-(u  —  v)  =  —  l,  2w  =  — 1,  or  u  = 


(1) 


Substituting  value  of  w  in  (1),  -  +  3  ^2  _j.  2  ^;4  -  97. 

0 


Solving  this, 


^2  =  25  Qj,  _31.  ^^^  ^,::3i5>  ^j.  _j_ 
4  4  2 


Then,  a;  =  w  +  v  =  -i±-,  or 
2      2 


and. 


,^^_,  =  _1^|,  or-lT 


2 
'^^31 


^  =  2,  -3,  or 


-  3,  2,  or 


The  solution  is   x  =  2,  ^ 


y  = 


31 


1- 


31 


3,   y=.2;    x  = 

_  _  1  +  v/^3l 


V-31 

2 

-i±V- 

-31 

2 

-itV: 

-31 

2 

-l+x/: 

-31 

MISCELLANEOUS  AND  REVIEW  EXAMPLES 

EXERCISE  145 

Solve  the  following  equations  : 

r2a;2/  +  ic  =  -36. 
"    [xy  —  Sy  =  —,5.  , 


2. 

"XL 

3. 
4. 


xy  =  6. 
if  +  V  =  \\K 

a^  —  Sxy  —  4:y^  =  0, 
Sx  —  5y  =46. 

rar^-22/2  +  3a5  =  -8. 


1 

-^ 

=  74 

1 

1 

=  12 

.  aj 

V 

2x_       _11 


I.  \y 


9. 


10 


x-y  =  l. 

x--  =  - 

y 


2     36 
2/4--  =  — 

=  17. 


p^  +  2/'  =  17 
I    a;-y  =  3. 

11.  P 


4d-hA;-3dA;  =  -6. 
5A;  +  2(?A;  =  10. 


296 


ALGEBRA 


12. 


13. 


iT     xy     y^ 
X      y 


I  -^  +  -\^y  ==  5. 


/^ 


14. 


15. 


16. 


llx^~xy  —  y^  =  45. 
7x^  +  3xy-2y^  =  20. 
a^+4:Xy  =  lS, 
[2xy  +  9y^  =  S7. 
a^y^  -  24:  xy-^  95  =  0, 

Sx-2y=-lS. 


(3x'-5xy  =  2a'-\-13ub-~7b\ 
1  x  +  y  =  3{a-b). 


18. 


19. 


20. 


3x-i-2y     3x-2y^41 

3x-2y     3x  +  2y     20* 

Sy^  +  3x'  =  29. 


21 


■I 


22. 


—  =  6a2. 
xy 

.x-\-y  =  5  axy. 

i+J=-^^^- 

ix^  y 

'  6^  +  9^2 _^4e  =  9. 

e^4-2^  =  -2. 

i»3  4-2/'  =  2a^.4-24a. 

.a^^  +  a;/  =  2a3-8a. 

26. 


27. 


28. 


23    I     -V2x^-9  =  3y  +  6. 

1  V«^-17/  =  a;2-5. 


'  3x^—xy—  xz  =  4:. 
5x-2y  =  l. 
4a;H-32  =  —  5. 
jc2/^  =  56. 
+  2/  =  -l. 


30 


31 


■  { 

■  1 

■  I 


r- 


25. 


«2     ^_19 
^"^«~"6"* 

1  +  1=1. 

a;      !/      6 

3a^  +  3/  =  10ci^. 

1  +  1  =  1 
X     y     3 

^y  4-  2/^aj  =  42. 

1+1=1. 

£c     2/     6 

5g2  +  gs-3s2  =  27. 
4g2_4^g_^352^72. 

y^-\-4:Xy  —  3y  =  4:2. 
2y^  —  xy  +  5y  =  —  10. 

16a^y^-104:xy  =  -105. 
x-y=z-2. 


32^  {^^  +  ^¥  +  2/^  =  481. 
^  '   \     x"- 


I        x- 


33. 


aj2/  +  /  =  37. 

9a^-13ajy-3a;=-123. 

a^2/+4/+22/=125. 


*  Divide  the  first  equation  by  the  second. 


SIMULTANEOUS  QUADRATIC  EQUATIONS        297 


34. 


35. 


36 


37 


38 


7f      xy     Y 

^=-12. 


xy     f 

.    I      x  —  3y  +  z  =  17. 
[     x-\-y  —  3z  =  13. 

1  x-y  =  j\. 

+  y)  +  xy  =11. 

yy+a^y^  =  61. 


39. 


40. 


41, 


42. 


43. 


xy{x  —  y)  =  —  20. 


aj2  +  2/^  =  a;2/  +  l. 
+  2/^  =  a^2/'  +  l. 


r  05^  +  2/5  =  211. 
1    a^  +  2/  =  l. 

0^2^  —  a?  =  —  6. 

ar^  =  -72. 


11      1  ^a'+a^h''+h\ 
Q?     xy    y^  a^W 


i_-l_Ll=^ 


■a262+64 


a^    fl??/    2/^ 


a^^^ 


PROBLEMS   IN   PHYSICS 

1.   From  the  equations  v  =  gt  and  S  =  ^gf,  find  v  in  terms 


of  S  and  gr. 


^ 


find    H   in 


2.  From   the   equations    C  —  —   and   ^(7 
terms  of  C,  i^,  and  t. 

3.  From  the  equations  E  =  -F'aS,  -F  =  ma,  S  =  \  af,  and  'y  =  at, 
find  ^  in  terms  of  m  and  v. 

313.  Problems  involving  Simultaneous  Equations  of  Higher 
Degree. 

In  solving  problems  which  involve  simultaneous  equations 
of  higher  degree,  only  those  solutions  should  be  retained  which 
satisfy  the  conditions  of  the  problem.     (Compare  §  176.) 

EXERCISE  146  '^ 

^  1.   The  difference  of  the  squares  of  two  numbers  is  56,  and 
the  difference  of  the  numbers  is  ^  their  sum.    Find  the  numbers. 
2.   The  sum  of  the  squares  of  two  numbers  is  61,  and  the 
product  of  their  squares  is  900.     Find  the  numbers. 


298  ALGEBRA 

3.  The  product  of  the  sum  of  two  numbers  by  the  smaller 
is  21,  and  the  product  of  their  difference  by  the  greater  is  4. 
Find  the  numbers. 

}  4.  The  sum  of  the  cubes  of  two  numbers  is  224 ;  and  if  the 
product  of  the  numbers  be  subtracted  from  the  sum  of  their 
squares,  the  remainder  is  28.     Find  the  numbers. 

5.  Two  numbers  are  expressed  by  the  same  two  digits  in 
reverse  order.  The  sum  of  the  numbers  equals  the  square  of 
the  sum  of  the  digits,  and  the  difference  of  the  numbers  equals 
5  times  the  square  of  the  smaller  digit.     Find  the  numbers. 

6.  The  square  of  the  sum  of  two  numbers  exceeds  their 
product  by  84 ;  and  the  sum  of  the  numbers,  plus  the  square 
root  of  their  product,  equals  14.     Find  the  numbers. 

7.  The  difference  of  the  cubes  of  two  numbers  is  342 ;  and 
if  the  product  of  the  numbers  be  multiplied  by  their  difference, 
the  result  is  42.     Find  the  numbers. 

8.  A  party  at  a  hotel  spent  a  certain  sum.  Had  there  been 
5  more,  and  each  had  spent  50  cents  less,  the  bill  would  have 
been  $24.75.  Had  there  been  3  fewer,  and  each  had  spent 
50  cents  more,  the  bill  would  have  been  $  9.75.  How  many 
were  there,  and  what  did  each  spend  ? 

9.  The  simple  interest  of  $700,  for  a  certain  number  of 
years,  at  a  certain  rate,  is  $  182.  If  the  time  were  4  years  less, 
and  the  rate  1^%  more,  the  interest  would  be  $  133.  Find  the 
time  and  the  rate. 

10.  If  the  digits  of  a  number  of  two  figures  be  inverted,  the 
quotient  of  this  number  by  the  given  number  is  If,  and  their 
product  1008.     Find  the  number. 

11.  The  square  of  the  smaller  of  two  numbers,  added  to  twice 
their  product,  gives  7  times  the  smaller  number ;  and  the  square 
of  the  greater  exceeds  the  product  of  the  numbers  by  6  times 
the  smaller  number.     Find  the  numbers. 

12.  A  rectangular  piece  of  cloth,  when  wet,  shrinks  one-sixth 
in  its  length,  and  one-twelfth  in  its  width.  If  the  area  is 
diminished  by  12|  square  feet,  and  the  length  of  the  four  sides 
by  6|^  feet,  find  the  original  dimensions. 


SIMULTANEOUS   QUADRATIC   EQUATIONS        299 

13.  A  and  B  travel  from  P  to  Q,  14  miles,  at  uniform  rates, 
B  taking  one-third  of  an  hour  longer  than  A.  to  perform  the 
journey.  On  the  return,  each  travels  one  mile  an  hour  faster, 
and  B  now  takes  one-fourth  of  an  hour  longer  than  A.  Find 
their  rates  of  travelling. 

14.  A  and  B  run  a  race  of  two  miles,  B  winning  by  two 
minutes.  A  now  inct'eases  his  speed  by  two  miles  an  hour, 
and  B  diminishes  his  by  the  same  amount,  and  A  wins  by  two 
minutes.     Find  their  original  rates. 

15.  A  man  ascends  the  last  half  of  a  mountain  at  a  rate  one- 
half  mile  an  hour  less  than  his  rate  during  the  first  half,  and 
reaches  the  top  in  3J  hours.  On  the  descent,  his  rate  is  one 
mile  an  hour  greater  than  during  the  first  half  of  the  ascent, 
and  he  accomplishes  it  in  2i  hours.  Find  the  distance  to  the 
top,  and  his  rate  during  the  first  half  of  the  ascent. 

16.  The  square  of  the  second  digit  of  a  number  of  three 
digits  exceeds  twice  the  sum  of  the  first  and  third  by  3.  The 
sum  of  the  first  and  second  digits  exceeds  4  times  the  third  by 
1 ;  and  if  495  be  subtracted  from  the  number,  the  digits  will  be 
inverted.     Find  the  number. 

17.  A  ship  has  provisions  for  36  days.  If  the  crew  were  16 
greater,  and  the  daily  ration  one-half  pound  less,  the  provisions 
would  last  30  days ;  if  the  crew  were  2  fewer,  and  the  daily 
ration  one  pound  greater,  they  would  last  24  days.  Find  the 
number  of  men,  and  the  daily  ration. 

18.  A  man  lends  f  2100  in  two  amounts,  at  different  rates  of 
interest,  and  the  two  sums  produce  equal  returns.  If  the  first 
portion  had  been  loaned  at  the  second  rate,  it  would  have  pro- 
duced $48;  and  if  the  second  portion  had  been  loaned  at  the 
first  rate,  it  would  have  produced  $  27.     Find  the  rates. 

19.  A  can  do  a  piece  of  work  in  2  hours  less  time  than  B ; 
and  together  they  can  do  the  work  in  li  hours  less  time  than 
A  alone.     How  long  does  each  alone  take  to  do  the  work  ? 


300 


ALGEBRA 


GRAPHICAL  REPRESENTATION  OF  SIMULTANEOUS  QUAD- 
RATIC EQUATIONS  WITH  TWO  UNKNOWN  NUMBERS 

314.   1.    Consider  the  equation  x^-\-y'^  =  25. 

This  means  that,  for  any  point  on  the  graph,  the  square  of  the  abscissa, 
plus  the  square  of  the  ordinate,  equals  25. 

But  the  square  of  the  abscissa  of  any  point, 
plus  the  square  of  the  ordinate,  equals  the 
square  of  the  distance  of  the  point  from  the 
origin  ;  for  the  distance  is  the  hypotenuse  of 
a  right  triangle,  whose  other  two  sides  are  the 
abscissa  and  ordinate. 

Then,  the  square  of  the  distance  from  0  of 
any  point  on  the  graph  is  25  ;  or,  the  distance 
from  0  of  any  point  on  the  graph  is  5. 

Thus,  the  graph  is  a  circle  of  radius  5,  having  its  centre  at  0. 

(The  graph  of  any  equation  of  the  form  x"^  -{■  y'^  =  a  i^  di.  circle.) 


2.  Consider  the  equation  2/^  =  4  a?  +  4. 

nx  =  0,       2/2  =  4,  or  ?/=±2.  {A,B) 

ft  _ 

If  cc  =  1,      ?/2  =  8,  or  y  =  ±  2  V2.     (C,  D) 

If  x  =  -\,  y  =0.  (E) 

etc. 

The  graph  extends  indefinitely  to  the  right  of 
YT. 

If  X  is  negative,  and  <  —  1,  y^  is  negative,  and 
therefore  y  imaginary  ;  then,  no  part  of  the  graph  lies  to  the  left  of  E. 

(The  graph  of  Ex.  2  is  2^ parabola;  as  also  is  the  graph  of  any  equation 
of  the  form  y"^  =  ax  or  y^  =  ax  -{-  b. 

The  graphs  of  §§  303  and  305  are  parabolas.) 

3.  Consider  the  equation  a;^  +  4  ?/^  =  4. 
In  this  case,  it  is  convenient  to  first 

locate  the  points  where  the  graph  inter- 
sects the  axes. 


If  i/  =  0. 

x2  =  4, 

or 

X- 

±2. 

{A,  A') 

If  X  =  0, 

41/2  =  4, 

or 

y  = 

±1. 

(B,  B') 

Putting  X 

=  ±1,  4 

[y2 

=  3 

2/2  = 

J,  or  2/: 

or2/  =  ±^.     (C,  A  CD') 


SIMULTANEOUS  QUADRATIC   EQUATIONS        301 


If  X  has  any  value  >2,  or  <  —  2,  y''^  is  negative,  and  y  imaginary  ;  then, 
no  part  of  the  graph  Ues  to  the  right  of  A,  or  left  of  AK 

If  y  has  any  value  >  1,  or  <  -  1,  x^  is  negative,  and  x  imaginary  ;  then, 
no  part  of  the  graph  lies  above  B^  or  below  B'. 

(The  graph  of  Ex.  3  is  an  ellipse  ;  as  also  is  the  graph  of  any  equation 
of  the  form  ax^  -^  by^  =  c.) 


4.   Consider  the  equation  a^  - 
Here,  x"^  -  l  =  2y^,  oi  y^  =  ^^~^ 


2  7/^-1 


2 

If  x=±l,  2/2=0,  or  y=0.      {A,  A') 

If  X  has  any  value  between  1  and 
—  1,  1/2  is  negative,  and  y  imaginary. 

Then,  no  part  of  the  graph  lies  be- 
tween A  and  A'. 

If  x=±2,  2/2  =|,  or  y  =  ±-\^- 

The  graph  has  two  branches,  BAC  and  B'A'C,  each  of  which  extends 
to  an  indefinitely  great  distance  from  O. 

(The  graph  of  Ex.  4  is  a  hyperbola  ;  as  also  is  the  graph  of  any  equation 
of  the  form  ax^  —  by^  =  c,  or  xy  —  a.) 


(B,  C,  B',  C) 


5.  4a^4-9/=36. 

6.  4ic2_2^2^_4 


EXERCISE  147 

Plot  the  graphs  of  the  following : 

1.  xy  =  -Q.  3.   a;2^2/'  =  4. 

2.  x'^^y.  4.   2/2  =  5a;-l. 

315.   Graphical  Representation   of  Solutions   of  Simultaneous 

Quadratic  Equations. 

r  2/^  =  4  ic. 

1.   Consider  the  equations  i  • 

ydx  —  y  =  o. 

The  graph  of  ?/2  =  4  x  is  the  parabola  A  OB. 

The  graph  oiZx  —  y  =  b  is  the  straight  line  AB^ 
intersecting  the  parabola  at  the  points  A  and  JB, 
respectively. 

To  find  the  co-ordinates  of  A  and  J5,  we  proceed 
as  in  §  184  ;  that  is,  we  solve  the  given  equations. 


The  solution  is  x 
(§  309). 


25 
1,  ^  =  -2;  or,  x  =  —,  y  = 


10 


302 


ALGEBRA 


It  may  be  verified  in  the  figure  that  these  are  the  co-ordinates  of  A  and 
B^  respectively. 

Hence,  if  any  two  graphs  intersect,  the  co-ordinates  of  any  point 
of  intersection  form  a  solution  of  the  set  of  equations  represented 
by  the  graphs. 

2.   Consider  the  equations 

(x^  +  f  =  17. 
[        xy  =  4:. 

The  graph  of  cc^  +  ^/^  =  17  is  the  circle 
AD,  whose  centre  is  at  0,  and  radius  \/l7. 

The  graph  of  xy  =  4:  is  a  hyperbola, 
having  its  branches  in  Ihe  angles  XOY 
and  X'OT,  respectively,  and  intersecting 
the  circle  at  the  points  A  and  B  in  angle 
XO  F,  and  at  the  points  C  and  D  in  angle  X'  0  T'. 

The  solution  of  the  given  equations  is  (§310), 

X  =  4:,  y  =  l  ;  cc=l,  ?/  =  4;  ic  =  —  1,  ?/=  —  4;  and  5C  =  —  4,  y  =  —l. 

It  may  be  verified  in  the  figure  that  these  are  the  co-ordinates  of  A,  B, 
C,  and  Z>,  respectively. 

EXERCISE  148 


Find  the  graphs  of  the  following  sets  of  equations,  and  in 
each  case  verify  the  principle  of  §  315 : 

9aj2-f-/=148. 
xy=-S. 
2/2  =  29. 
xy  =  10. 


raj2  4-4?/2  =  4. 

I       x  —  y  =  l. 
J    r    x'-4.y=-7. 
''  l2ic-f-3^=4. 


I- 1'* 


'  1  3x2-4/= -24. 
g    r     0.2.^2/^  =  13. 
*   [4:x-9v  =  6. 


316.   1.   Consider  the  equations 

rx2-f42/2  =  4.  (1) 

[2x  +  3y  =  -5.  (2) 

The  graph  of  a;^^  4  2/2—4  jg  h^q  ellipse 
AB. 

The  graph   of   2x  +  3?/  =  —  5  is  the 
straight  line  CD. 


SIMULTANEOUS  QUADRATIC   EQUATIONS        303 


To  solve  the  given  equations,  we  have,  from  (2),  x= ^ • 

Substituting  in  (1),     ^^'  +  '^^  +  ^^  +  4y^  =  ^. 

Then,  25  ^/^  +  30  y  +  9  =  0,  or  (5  ?/  +  3)(5  y  +  3)  =  0. 

o  g 

This  equation  has  equal  roots;  the  only  value  of  i/  is  —  - ;  and  x  =  --• 

5  o 

The  line  has  but  one  point  in  common  with  the  ellipse,  and  is  tangent 
to  it. 

Then,  if  the  equation  obtained  by  eliminating  one  of  the  un- 
known numbers  has  equal  roots,  the  graphs  are  tangent  to  each  other. 

2.   Consider  the  equations 

f9a;2-2/^  =  -9.  (1)  \l 

I    x-2y  =  ~2.  \\y 

The  graph  of  9  ic^  —  ?/2  =  —  9  is  a  hyperbola, 
having  its  branches  above  and  below  XX',  re- 
spectively. 

The  graph  of  x  —  2y  =  —  2  is  the  straight  line 
AB. 

To  solve  the  given  equations,  we  substitute 
a;  =  2?/-2  in  (1). 

Then,     9(4  2/2_  8  ?/ +  4)-?/2  =  -9, 

or  35  ^2  _  72  y  +  45  =  0. 

This  equation  has  imaginanj  roots,  which  shows  that  the  line  does  not 
intersect  the  hyperbola. 

In  general,  if  the  equation  obtained  by  eliminating  one  of  the 
unkiiown  numbers  has  imaginary  roots,  the  graphs  do  not  intersect. 


Exercise  i49 

Find  the  graphs  of  the  following  sets  of  equations,  and  in 
each  case  verify  the  principles  of  §  316 : 

\sx-y  =  ^.  '   I 


x-'-y- 
5  if  —  4  ?/  = 


{f-Zx  =  -S.  I      x'^  +  f=\. 

'  I  x  +  2y^-2.  '   [2y'-3x  =  5. 


304  ALGEBRA 


XXIII.  VARIABLES  AND  LIMITS 

317.  A  variable  number,  or  simply  a  variable,  is  a  number 
which  may  assume,  under  the  conditions  imposed  upon  it,  an 
indefinitely  great  number  of  different  values. 

A  constant  is  a  number  which  remains  unchanged  throughout 
the  same  discussion. 

318.  A  limit  of  a  variable  is  a  constant  number,  the  differ- 
ence between  which  and  the  variable  may  be  made  less  than 
any  assigned  number,  however  small. 

Suppose,  for  example,  that  a  point  moves  from  A  towards 
B  under  the  condition  that  it  shall  move,  during  successive 
equal  intervals  of  time,  first 

from  A  to  C,  half-way  between     -f ^ L___L__f 

A  and  B',  then  to  D,  half-way 


between  C  and  B ;  then  to  E,  half-way  between  D  and  B ;  and 
so  on  indefinitely. 

In  this  case,  the  distance  between  the  moving  point  and  B 
can  be  made  less  than  any  assigned  number,  however  small. 

Hence,  the  distance  from  A  to  the  moving  point  is  a  vari- 
able which  approaches  the  constant  value  AB  as  a  limit. 

Again,  the  distance  from  the  moving  point  to  5  is  a  variable 
which  approaches  the  limit  0. 

319.   Interpretation  of  -• 

Consider  the  series  of  fractions  -,  — ,  -— -,  —-—,  •••. 

3'  .3'  .03'  .003' 

Here  each  denominator  after  the  first  is  one-tenth  of  the 
preceding  denominator. 

It  is  evident  that,  by  sufficiently  continuing  the  series,  the 
denominator  may  be  made  less  than  any  assigned  number, 
however  small,  and  the  value  of  the  fraction  greater  than  any 
assigned  number,  however  great. 


VARIABLES  AND  LIMITS  305 

In  other  words, 


If  the  numerator  of  a  fraction  remains  constant,  while  the 
denominator  approaches  the  limit  0,  the  value  of  the  fraction 
increases  without  limit. 

It  is  customary  to  express  this  principle  as  follows : 
a 

The  symbol  oo  is  called  Infinity ;  it  simply  stands  for  that  which  is 
greater  than  any  number,  however  great,  and  has  no  fixed  value. 

320.   Interpretation  of  —  • 

GO 

Consider  the  series  of  fractions 


3'  30'  300'  3000' 

Here  each  denominator  after  the  first  is  ten  times  the  pre- 
ceding denominator. 

It  is  evident  that,  by  sufficiently  continuing  the  series,  the 
denominator  may  be  made  greater  than  any  assigned  number, 
however  great,  and  the  value  of  the  fraction  less  than  any 
assigned  number,  however  small. 

In  other  words, 

If  the  numerator  of  a  fraction  remains  constant,  while  the 
denominator  increases  without  limit,  the  value  of  the  fraction 
approaches  the  limit  0. 

It  is  customary  to  express  this  principle  as  follows : 

-^  =  0. 

00 

321.   No  literal  meaning  can  be  attached  to  such  results  as 

a  a      p, 

-  =  oo,  or   -  =  0; 

0  00 

for  there  can  be  no  such  thing  as  division  unless  the  divisor  is 
2ifi7iite  number. 

If  such  forms  occur  in  mathematical  investigations,  they 
must  be  interpreted  as  indicated  in  §§  319  and  320.  (Com- 
pare §  420.) 


306  ALGEBRA 


THE  PROBLEM   OF  THE  COURIERS 

322.  The  following  discussion  will  further  illustrate  the 
form  - ,  besides  furnishing  an  interpretation  of  the  form  -  • 

The  Problem  of  the  Couriers. 

Two  couriers,  A  and  B,  are  travelling  along  the  same  road  in. 
the  same  direction,  RW,  at  the  rates  of  m  and  n  miles  an  hour, 
respectively.  If  at  any  time,  say  12  o'clock,  A  is  at  P,  and  B 
is  a  miles  beyond  him  at  Q,  after  how  many  hours,  and  how 
many  miles  beyond  P,  are  they  together  ? 

R  P  Q  R- 

I J I I 


Let  A  and  B  meet  x  hours  after  12  o'clock,  and  y  miles 
beyond  P. 

They  will  then  meet  y  —  a  miles  beyond  Q. 

Since  A  travels  mx  miles,  and  B  nx  miles,  in  x  hours,  we 

^^^®  f        y  =  mx, 

y  —  a  =  nx. 

Solving  these  equations,  we  obtain 


m  —  7i 
We  will  now  discuss  these  results  under  different  hypotheses. 

1.  m>n. 

In  this  case,  the  values  of  x  and  y  are  positive. 

This  means  that  the  couriers  meet  at  some  time  after  12,  at 
some  point  to  the  right  of  P. 

This  agrees  with  the  hypothesis  made ;  for  if  m  is  greater 
than  n,  A  is  travelling  faster  than  B ;  and  he  must  overtake 
him  at  some  point  beyond  their  positions  at  12  o'clock. 

2.  m<n. 

In  this  case,  the  values  of  x  and  y  are  negative. 
This  means  that  the  couriers  met  at  some  time  before  12,  at 
some  point  to  the  left  of  P.     (Compare  §  16.) 


VARIABLES  AND  LIMITS  307 

This  agrees  with  the  hypothesis  made ;  for  if  m  is  less  than 
n,  A  is  travelling  more  slowly  than  B ;  and  they  must  have 
been  together  before  12  o'clock,  and  before  they  could  have 
advanced  as  far  as  P. 

3.  a  =  0,  and  m>n  or  m<n. 
In  this  case,  a?  =  0  and  y  =  0. 

This  means  that  the  travellers  are  together  at  12  o'clock,  at 
the  point  P. 

This  agrees  with  the  hypothesis  made ;  for  if  a  =  0,  and  m 
and  n  are  unequal,  the  couriers  are  together  at  12  o'clock,  and 
are  travelling  at  unequal  rates ;  and  they  could  not  have  been 
together  before  12,  and  will  not  be  together  afterwards. 

4.  m  =  71,  and  a  not  equal  to  0. 

In  this  case,  the  values  of  x  and.  2/  take  the  forms  -  and  — , 
respectively. 

li  m  —  n  approaches  the  limit  0,  the  values  of  x  and  y  increase 
without  limit  (§  319) ;  hence,  if  m  =  n,  no  fixed  values  can  be 
assigned  to  x  and  y,  and  the  problem  is  impossible. 

In  this  case,  the  result  in  the  form  -  indicates  that  the  given 
problem  is  impossible. 

This  agrees  with  the  hypothesis  made ;  for  if  m  =  n,  and  a 
is  not  zero,  the  couriers  are  a  miles  apart  at  12  o'clock,  and  are 
travelling  at  the  same  rate ;  and  they  never  could  have  been, 
and  never  will  be  together. 

5.   m  =  n,  and  a  =  0. 

In  this  case,  the  values  of  x  and  y  take  the  form  — 

^  0 

If  a  =  0,  and  m  =  n,  the  couriers  are  together  at  12  o'clock, 
and  travelling  at  the  same  rate. 

Hence,  they  always  have  been,  and  always  will  be,  together. 

In  this  case,  the  number  of  solutions  is  indefinitely  great ; 
for  any  value  of  x  whatever,  together  with  the  corresponding 
value  of  y,  will  satisfy  the  given  conditions. 

In  this  case,  the  result  in  the  form  -  indicates  that  the  number 
of  solutions  is  indejinitely  great. 


808  ,  ALGEBRA 

XXIV.    INDETERMINATE  EQUATIONS 

323.  It  was  shown,  in  §  163,  that  a  single  equation  involving 
two  or  more  unknown  numbers  is  satisfied  by  an  indefinitely 
great  number  of  sets  of  values  of  these  numbers. 

If,  however,  the  unknown  numbers  are  required  to  satisfy 
other  conditions,  the  number  of  solutions  may  be  finite. 

We  shall  consider  in  the  present  chapter  the  solution  of 
indeterminate  linear  equations,  in  which  the  unknown  numbers 
are  restricted  to  positive  integral  values. 

324.  Solution  of  Indeterminate  Linear  Equations  in  Positive 
Integers. 

1.    Solve  7  ic  +  5  2/  =  118  in  positive  integers. 
Dividing  by  6,  the  smaller  of  the  two  coefficients, 

aj  +  2^+2/  =  23  +  |;  or,  ^J^=2Z-x-y. 

0  0  0 

Since,  by  the  conditions  of  the  problem,  x  and  y  must  be  positive  inte- 

2  X 3 

gers,  must  be  an  integer. 

5 

Let  this  integer  be  represented  by  p. 

Then,  ^^~^  =  p,  or2a;-3  =  5p.  .  (1) 

5 

Dividing  (1)  by  2,  x- 1 -i=  2i)  H- 1 ;    or,  x-l-2p  =  ^^. 

Since  x  and  p  are  integers,  ic  —  1— 2j5isan  integer ;   and  therefore 

r)  4-  1 

^^— - —  must  be  an  integer. 

z 

Let  this  integer  be  represented  by  q. 

Then,  -2_+i  =  gr,  or  p  =  2q  -h 

Substituting  in  (1),  2  ic  -  3  =  10  g'  -  5. 

Whence,  x-hq-\.  (2) 


INDETERMINATE   EQUATIONS  309 

Substituting  this  value  in  the  given  equation, 

35  g  -  7  +  5  y  =  118  ;  or,  y  =  25  -  7  g.  (3) 

Equations  (2)  and  (3)  form  the  general  solution  in  integers  of  the 
given  equation. 

By  giving  to  q  the  value  zero,  or  any  positive  or  negative  integer,  we 
shall  obtain  sets  of  integral  values  of  x  and  y  which  satisfy  the  given 
equation. 

If  q  is  zero,  or  any  negative  integer,  x  will  be  negative. 

If  q  is  any  positive  integer  >3,  ?/  will  be  negative. 

Hence,  the  only  positive  integral  values  of  x  and  y  which  satisfy  the 
given  equation  are  those  obtained  from  the  values  1,  2,  3  of  g'. 

That  is,  a;  =  4,  y  =  18  ;  x  =  9,  y  =  11 ;  and  a:  =  14,  ?/  =  4. 

2.  Solve  5  a;  —  7  2/  =  11  in  least  positive  integers. 
Dividing  by  5,  the  coefficient  of  smaller  absolute  value, 

5  5  6 

2  w  4-  1 
Then,     ^  ^     must  be  an  integer. 
5 

Let  ?^-±^  =i? ;  or,  2  ?/ +  1  =  5p. 
5 

Dividing  by  2,  i/+i  =  2p+^;  or,  y-2j)  =  ^-=^. 

Then,  ^  "~     must  be  an  integer. 

Let  ^  ~     =  g  ;  or,  jp  =  2  g  +  1. 

Then,  y  =  5p-1^10g  +  5-l^^^^^_ 

Then,  from  the  given  equation,  x  =    ^  "j" —  =  7  g  +  5. 

5 

The  solution  in  least  positive  integers  is  when  ^-=0  ;  that  is,  x=6,y=2. 

3.  In  how  many  ways  can  the  sum  of  f  15  be  paid  with 
dollars,  half-dollars,  and  dimes,  the  number  of  dimes  being 
equal  to  the  number  of  dollars  and  half-dollars  together  ? 

Let  X  =  number  of  dollars, 

y  =  number  of  half-dollars, 

and  z  =  number  of  dimes. 


Adding, 

llx  +  6ij  +  z  =  150  +  z, 

or, 

nx  +  6y  =  150. 

Dividing  by  6, 

x  +  ^+y  =  25. 

310  ALGEBRA 

riOx+52/+2r=150, 
By  the  conditions,    1 

I  a:  +  ?/  =  0.  (1) 


(2) 


5  X 
Then,  —  must  be  an  integer ;  or,  x  must  be  a  multiple  of  6. 

6 
Let  X  =  6p,  where  p  is  an  integer. 

Substitute  in  (2),  66p  +  6y  =  150,  or  ?/  =  25  -  lip. 

Substitute  in  (1),  s  =  6p -{-26  -  Up  =  25  -  bp. 

The  only  positive  integral  solutions  are  when  j?  =  1  or  2. 
Then,  the  number  of  ways  is  two  ;  either  6  dollars,  14  half-dollars,  and 
20  dimes  ;  or,  12  dollars,  3  half-dollars,  and  15  dimes. 

EXERCISE  150 
Solve  the  following  in  positive  integers : 

1.  Sx-\-5y  =  29.  7.   2Sx-\-9y  =  Wl. 

2.  7x-\-2y  =  S9.  8.   8 x  +  71 2/ =  1933. 

V   3.  6x-\-29y  =  2T4:,  ^  f  8aj-ll2/  +  2^  =  10. 

4.  4a;  +  3l2/  =  473.  '  I      2x-9 y-{-z  =  -S. 

5.  42i»  +  ll2/  =  664.  ^^  r3a^-32/H-7;2  =  101. 

6.  10a;  +  72/  =  297.  *  [4.x-^2y -3z  =  5, 

Solve  the  following  in  least  positive  integers : 

11.   6x-7y  =  4:.  14.   8a;-3l2/  =  10. 

V^12.   5a;-8?/  =  17.  15.   SO  x  -  IS  y  =.  115. 

13.   14aj-52/  =  64.  16.   15a!-38  2/  =  -47. 

"^    17.   In  how  many  different  ways  can  f  1.65  be  paid  with 
quarter-dollars  and  dimes  ? 

18.  In  how  many  different  ways  can  41  shillings  be  paid 
with  half-crowns,  worth  2^  shillings  each,  and  two-shilling 
pieces  ? 


INDETERMINATE   EQUATIONS  311 

J 

19.  Find  two  fractions  whose  denominators  are  5  and  7, 
respectively,  whose  numerators  are  the  smallest  possible  posi- 
tive integers,  and  whose  difference  is  ^|^. 

20.  In  how  many  different  ways  can  $7.15  be  paid  with 
fifty-cent,  twenty-five  cent,  and  twenty-cent  pieces,  so  that 
twice  the  number  of  fifty-cent  pieces,  plus  twice  the  number  of 
twenty-cent  pieces,  shall  exceed  Ihe  number  of  twenty-five  cent 
pieces  by  31  ? 

21.  A  farmer  purchased  a  certain  number  of  pigs,  sheep,  and 
calves  for  f  138.  The  pigs  cost  $  4  each,  the  sheep  f  7  each, 
and  the  calves  $9  each;  and  the  whole  number  of  animals 
purchased  was  23.     How  many  of  each  did  he  buy  ? 

22.  In  how  many  ways  can  $  10.00  be  paid  with  twenty-five 
cent,  twenty-cent,  and  five-cent  pieces,  so  that  3  times  the  num- 
ber of  twenty-five  cent  pieces,  plus  15  times  the  number  of 
twenty-cent  pieces,  shall  exceed  the  number  of  five-cent  pieces 
by  33  ? 


312  ALGEBRA 


XXV.    RATIO  AND  PROPORTION 


RATIO 

325.  The  Ratio  of  one  nuin'ber  a  to  another  number  &  is  the 
quotient  of  a  divided  by  h. 

Thus,  the  ratio  of  a  to  6  is  -  ;  it  is  also  expressed  a :  h, 

b 

In  the  ratio  a:b,  a  is  called  the  Jirst  term,  or  antecedent^  and 

h  the  second  term,  or  consequent. 

If  a  and  b  are  positive  numbers,  and  a>b,  -  is  called  a  ratio 

of  greater  inequality;  if  a  is  <b,  it  is  called  a  ratio  of  less 
inequality. 

326.  A  ratio  of  greater  inequality  is  decreased,  and  one  of  less 
iyiequality  is  increased,  by  adding  the  same  positive  number  to 
each  of  its  terms. 

Let  a  and  b  be  positive  numbers,  a  being  >  b,  and  x  a  positive 
number. 

Since  a>b,  ax> bx.  (§  195) 

Adding  ab  to  both  members  (§  192), 

ab  +  ax  >ab-\-  bx,  or  a(b  +  a?)  >  b{a  -f  x). 

Dividing  both  members  by  b(b  +  x),  we  have 

^>P^'  (§195) 

b      b-\-x 

•» 

In  like  manner,  if  a  is  <  &,  -  <  ^^-+^. 

'  b     b-^x 

PROPORTION 

327.  A  Proportion  is  an  equation  whose  members  are  equal 
ratios. 


RATIO   AND  PROPORTION  313 


Thus,  it  a:b  and  c :  d  are  equal  ratios 


is  a  proportion. 


a:o  =  c:d,  or  -  =  -. 
b     d' 


328.  In  the  proportion  a:b  =  c:d,  a  is  called  the  first  term, 
b  the  second,  c  the  third,  and  d  the  fourth. 

The  first  and  third  terms  of  a  proportion  are  called  the  ante- 
cedents, and  the  second  and  fourth  terms  the  consequeyits. 

The  first  and  fourth  terms  are  called  the  extremes,  and  the 
second  and  third  terms  the  means. 

329.  If  the  means  of  a  proportion  are  equal,  either  mean  is 
called  the  Mean  Proportional  between  the  first  and  last  terms, 
and  the  last  term  is  called  the  Third  Proportional  to  the  first 
and  second  terms. 

Thus,  in  the  proportion  a:b  =  b:c,bis  the  mean  proportional 
between  a  and  c,  and  c  is  the  third  proportional  to  a  and  b. 

The  Fourth  Proportional  to  three  numbers  is  the  fourth  term 
of  a  proportion  whose  first  three  terms  are  the  three  numbers 
taken  in  their  order. 

Thus,  in  the  proportion  a:b  =  c:d,  d  is  the  fourth  propor- 
tional to  a,  b,  and  c. 

330.  A  Continued  Proportion  is  a  series  of  equal  ratios,  in 
which  each  consequent  is  the  same  as  the  next  antecedent  j  as, 

a:b  =  b:c  =  c:d=:d:e. 

PROPERTIES  OF  PROPORTIONS 

331.  In  any  proportion,  the  product  of  the  extremes  is  equal  to 
the  product  of  the  means. 

Let  the  proportion  be     a:b  =  c:d. 


a_c 
b~d 


Then  by  §  327, 

Clearing  of  fractions,       ad  =  be. 


314  ALGEBRA 

332.  From  the  equation  ad  =  be  (§  331),  we  obtain 

be    7      ad  ad,        -,    ■,      be 

a  =  —y  b  =  — ,  c  =  — ,  and  a  =  — . 
deb  a 

That  is,  171  any  proportion,  either  extx^rne_eguols^thejgrg^^ 
of  the  means  divided  by  the  otlier^xtreme^i^giuA^UJxBi^mEaji^^^ 
tis-prodMe^of^the^sctrein^ 

333.  (Converse  of  §  331.)  If  the  product  of  tivo  numbers  be 
equal  to  the  product  of  two  others,  one  pair  may  be  made  the 
extremes,  and  the  other  pair  the  means,  of  a  proportion. 


Let 

ad  =  bc. 

Dividing  by  bd. 

ad  _hc      .  ci_c 
hd^ii  °^"  b~d 

Whence,  by  §[327j 

a:b  =  c:d. 

In  like  manner,  we 

may  prove  that 

a:c  =  b:d, 

c:d  =  a:b,  etc. 

334.  In  any  proportion,  the  terms  are  in  proportion  by  Alter- 
nation ;  that  is,  the  means  can  be  interchanged. 

Let  the  proportion  be    a  :  b  =  c  :  d. 

Then  by  §331,  ad  =  bc. 

Whence,  by  §  333,  a  :  c  =  6  :  d 

In  like  manner,  it  may  be  proved  that  the  extremes  can  be  inter- 
changed. '^'\o^^' 

335.  In  any  proportion,  the  terms  are  in  propoHion  by  Inver- 
sion ;  that  is,  th,e  second  term  is  to  the  first  as  the  fourth  term  is 
to  the  third. 

Let  the  proportion  be    a:b  =  c:  d. 
Then,  by  §331,  ad=bc. 

Whence,  by  §  333,  b:a  =  d:c. 


RATIO   AND  PROPORTION  315 

It  follows  from  §  335  that,  in  any  proportion,  the  means  can  be  written 
as  the  extremes,  and  the  extremes  as  the  means. 

336.  The  mean  proportional  between  two  numbers  is  equal  to 
the  square  root  of  their  product. 

Let  the  proportion  be    a:b  —  b:c. 

Then  by  §  331,         b"  =  ac,  and  b  =  Vac. 

337.  In  any  proportion,  the  terms  are  in  proportion  by  Com- 
position; that  is,  the  sum  of  the  first  two  terms  is  to  the  first 
term  as  the  sum  of  the  last  two  terms  is  to  the  third  term. 

Let  the  proportion  be    a:b  =  c:d. 

Then,  ad  =  be. 

Adding  each  member  of  the  equation  to  ac, 

ac -\- ad  =  ac -{-  be,  or  a(c  -f  c?)  =  c(a  +  b). 

Then  by  §333,  a-\-b  :  a  =  c-]-d:  c. 

We  may  also  prove        a  +  b  :  b  =  c  +  d  :  d. 

338.  In  any  proportion,  the  terms  are  in  proportion  by  Division; 
that  is,  the  difference  of  the  first  two  terms  is  to  the  first  term  as 
the  difference  of  the  last  tivo  terms  is  to  the  third  term. 

Let  the  proportion  be     a  :  b  =  c  :  d. 

Then,  ad  =  be. 

Subtracting  each  member  of  the  equation  from  ac, 

ac  —  ad  =  ac  —  be,  or  a(c  —  d)  —  c(a  —  b). 

Then,  a  —  b  :  a  =  c  —  d:c. 

We  may  also  prove         a  —  h  :  b  =  c  —  d  :  d. 

339.  hi  an?/  proportion,  the  terms  are  in  proportion  by  Com- 
position and  Division ;  that  is,  the  sum  of  the  first  two  terms  is 
to  their  difference  as  the  sum  of  the  last  tivo  terms  is  to  their 
difference. 

Let  the  proportion  be     a:b  =c:d. 


316  ALGEBRA 


Then  by  §  337,              ^^±^  =  ^-±-^.  (1) 

a            c  ^  ^ 

And  by  §  338,               ^^^  =  ^—^^  (2) 

a            c  ^ 


Dividing  (1)  by  (2), 


a  -\-b     c-\-  d 


a—b     c—d 
Whence,  a-\-b:a  —  b  =  c-\-d:c  —  d. 

340.  In  any  propoHion,  if  the  first  two  terms  be  multiplied  by 
any  number,  as  also  the  last  two,  the  resulting  numbers  will  be  in 
proportion. 

Let  the  proportion  be  -  =  - ;  then,  ^  =  ^. 

b      d  mb      nd 

(Either  m  or  n  may  be  unity  ;  that  is,  the  terms  of  either  ratio  may  be 
multiplied  without  multiplying  the  terms  of  the  other.) 

341.  I7i  any  proportion,  if  the  first  and  third  terms  be  multi- 
plied by  any  number,  as  also  the  second  and  fourth  terms,  the 
resulting  numbers  will  be  m  proportion. 

Let  the  proportion  be  -  =  - ;  then,    'B^^'^. 

b     d  nb      nd 

(Either  to  or  n  may  be  unity. ) 

342.  In  any  number  of  proportions,  the  products  of  the  cor- 
responding terms  are  in  proportion. 

Let  the  proportions  be         -  =  -,  and  -  =  2. 
^    ^  b     d'  f     h 

Multiplying,  ^x^  =  ^xf,  or^^  =  ^. 

b     f      d     h  bf     dh 

In  like  manner,  the  theorem  may  be  proved  for  any  number 
of  proportions. 

343.  In  any  proportion,  like  powers  or  like  roots  of  the  terms 
are  in  proportion. 


RATIO   AND  PROPORTION  317 

■  ■ »  I  I  n  '  I  r  I  r  ■        ■  '  ■  ■  '-^^  •  •■'  r  t 

Let  the  proportion  be         -  =  - ;  then,  —  =  — . 

T        Tl  'Vet  -\/g 

In  like  manner,  = • 

V6      ^/d 

344.  In  a  series  of  equal  ratios,  any  antecedent  is  to  its  con- 
sequent as  the  sum  of  all  the  ayitecedents  is  to  the  sum  of  all  the 
consequents. 

Let  a:b  =  c:d  =  e:f 

Then  by  §331,  ad  =  bc, 

and  af=be. 

Also,  ab  =  ba. 

Adding,  a(b-{-d-{-f)  =  b(a  +  c-\-e). 

Whence,*  a:b  =  a-\-c-\-e:b-\-d+f        (§333) 

In  like  manner,  the  theorem  may  be  proved  for  any  number 
of  equal  ratios. 

345.  If  three  numbers  are  in  continued  proportion,  the  first  is 
to  the  third  as  the  square  of  the  first  is  to  the  square  of  the  second. 

Let  the  proportion  be    a:  b  =  b  :c;  oi  -  =  -- 

b      c 

Then,  a     b_a     a      ,a^^ 

b     c      b      b         c      ¥ 

346.  If  four  numbers  are  in  continued  proportion,  the  first  is 
to  the  fourth  as  the  cube  of  the  first  is  to  the  cube  of  the  second. 

Let  the  proportion  be     a:b  =  b  :  c  =  c:  d;  ov   -  =  -  =  -. 

bed 

Then,  ^x-^^  =  ^X^X^,  or^  =  ^. 

b     Q     d     b     b     b         d     b^ 


318  ALGEBRA 

347.   Examples. 

1.    li  x:y  =  (x-i-zy:  (y  -{- zy,  prove  z  the  mean  proportional 
between  x  and  y. 

From  the  given  proportion,  by  §  331, 

yix  +  0)2  =  x(y  +  zy. 
Or  x"^  +  2  xyz  +  yz^  =  a;^^  _(_  2  aj?/^  4-  x^!^. 

Transposing,  x^y  —  xy^  =  xz'^  —  xjz'^. 

Dividing  by  aj  —  ?/,  xy  =  z^. 

Therefore,  z  is  the  mean  proportional  between  x  and  y  (§  336). 

The  theorem  of  §  339  saves  work  in  the  solution  of  a  certain 
class  of  fractional  equations. 

±   Sol^e  the  equation  ?^±|  =  ?-^^:i^. 
^  2x-3     26+a 

Regarding  this  as  a  proportion,  we  have  by  composition  and  division, 

4x        46       ^„2x         26    ^i,^„„^    „  3  6 

—  = ,  or  —  = ;  whence,  a;  = 

6       -2a'         3  a  a 

3.   Prove  that  if -  =  t-,  then  * 

h     d 

(^-b'':a'-3ab  =  c'-d':c'-3cd. 

Let  -  =  -  =  X,  whence,  a  =  6a; ;  then, 

6     cZ  c2_^ 

^2  _  ^2  ;,2a;2  _  52  y.2  _  1  (Z^  C2  -  (^2 


a2  -  3  «6      62x2  -  3  62x     x'^  -  3  x      «!  _  3c     c^  -3cd 

d^^    d 

Then,  a2  _  ^2  .  ^2  _  3  ^^5  =  c2  -  cZ2  :  c2  -  3  cd 


EXERCISE  151 

1.  Find  the  mean  proportional  between  18  and  32. 

2.  Find  the  third  term  of  a  proportion  whose  first,  second, 
and  fourth  terms  are  24,  32,  and  20,  respectively. 

3.  Find  the  third  proportional  to  J^  and  ^. 

4.  Find  the  mean  proportional  between  1^  and  24|-. 


RATIO   AND  PROPORTION  319 

5.  Find  the  fourth  proportional  to  4|,  5f,  and  1-f. 

6.  Find  the  third  proportional  to  ci^  +  8  and  a  +  2. 

7.  Find  the  mean  proportional  between 

—  and  -^ 

x  —  5  x-\-3 

Solve  the  following  equations  : 

Sx-S^2x-5         ^Q    ^^-2x-^^Sx±2^ 
'    3x  +  4:'~2x-\-7'  '   x^-2x-3     3x-2 


12. 


9    4a;  +  7^7a4-l  n.   «;-  + V3a;-1  ^a^- V2a;  +  1^ 

•    4a;-7      5a-3*  '   a^-V3a;-l      a^+V2a;  +  l 

a;4-y_a  — 6 

x  —  y     a  +  6  ♦ 

0?  —  g'^  _  6^  +  y  ^ 
a?  +  a^     h^  —  y 

13.  Find  two  numbers  in  the  ratio  4  to  3,  such  that  the  dif- 
ference of  their  squares  shall  be  112. 

14.  Find  two  numbers  such  that,  if  9  be  added  to  the  first, 
and  7  subtracted  from  the  second,  they  will  be  in  the  ratio  9:2; 
while  if  9  be  subtracted  from  the  first,  and  7  added  to  the  second, 
they  will  be  in  the  ratio  9 :  11. 

15.  Find  two  numbers  in  the  ratio  a :  &,  such  that,  if  each  be 
increased  by  c,  they  shall  be  in  the  ratio  m :  n. 

16.  Find  three  numbers  in  continued  proportion  whose  sum 
is  \7-,  such  that  the  quotient  of  the  first  by  the  second  shall 
be|. 

17.  What  number  must  be  added  to  each  of  the  numbers  a, 
6,  and  c,  so  that  the  resulting  numbers  shall  be  in  continued 
proportion  ? 

18.  Find  a  number  such  that,  if  it  be  subtracted  from  each 
term  of  the  ratio  8 : 5,  the  result  is  \\  of  what  it  would  have 
been  if  the  same  number  had  been  added  to  each  term. 


320  ALGEBRA 

19.  The  second  of  two  numbers  is  the  mean  proportional  be- 
tween the  other  two.  The  third  number  exceeds  the  sum  of 
the  other  two  by  20  ;  and  the  sum  of  the  first  and  third  exceeds 
three  times  the  second  by  4.     Find  the  numbers. 

20.  If  8a  —  56:7  a  —  46  =  86  —  5c:  7  6  —  4  c,  prove  c  the 
third  proportional  to  a  and  6. 

21.  If  ma-{-nb:  pa  —  qb  =  mc  -i-ndipc  —  qd,  prove  a:b  =  c:d. 


22.  If  x-{-y:y  -]-z  =  -Vxr  —  y^ :  V^/^  —  z^,  prove  y  the  mean 
proportional  between  x  and  z. 

23.  Given  (2  0^+2  ab)x-\-(a^+2b^)y=(a'-b')x-\-{2a^-\-b^)y, 
find  the  ratio  of  x  to  y. 

24.  If  4  silver  coins  and  11  copper  coins  are  worth  as  much 
as  2  gold  coins,  and  5  silver  coins  and  19  copper  coins  as  much 
as  3  gold  coins,  find  the  ratio  of  the  value  of  a  gold  coin,  and 
the  value  of  a  silver  coin,  to  the  value  of  a  copper  coin. 

TO  a     c 
6      a 

25.  3a4-46:3a-46  =  3c  +  4d:3c-4d 

26.  d'-5ab:2ab-\-7b^  =  c'-5cd:2cd  +  7d\ 

27.  a'  +  6ab':a'b-Sb'=c'-{-6cd':c'd-Sd^ 

28.  Each  of  two  vessels  contains  a  mixture  of  wine  and 
water.  A  mixture  consisting  of  equal  measures  from  the  two 
vessels  is  composed  of  wine  and  water  in  the  ratio  3:4;  another 
mixture  consisting  of  2  measures  from  the  first  and  3  measures 
from  the  second,  is  composed  of  wine  and  water  in  the  ratio 
2 : 3.     Find  the  ratio  of  wine  to  water  in  each  vessel. 


VARIATION  321 


J^XXVI.    VARIATION 

348.  One  variable  number  (§  317)  is  said  to  vary  directly  as 
another  when  the  ratio  of  any  two  values  of  the  first  equals 
the  ratio  of  the  corresponding  values  of  the  second. 

It  is  usual  to  omit  tlie  word  "directly,"  and  simply  say  that  one 
number  varies  as  another. 

Thus,  if  a  workman  receives  a  fixed  number  of  dollars  per 
diem,  the  number  of  dollars  received  in  m  days  will  be  to  the 
number  received  in  n  days  as  m  is  to  n. 

Then,  the  ratio  of  any  two  numbers  of  dollars  received  equals 
the  ratio  of  the  corresponding  numbers  of  days  worked. 

Hence,  the  number  of  dollars  which  the  workman  receives 
varies  as  the  number  of  days  during  which  he  works. 

349.  The  symbol  oc  is  read  ^^  varies  as^' ;  thus,  accb  is  read 
"  a  varies  as  6." 

350.  One  variable  number  is  said  to  vary  inversely  as  another 
when  the  first  varies  directly  as  the  reciprocal  of  the  second. 

Thus,  the  number  of  hours  in  which  a  railway  train  will 
traverse  a  fixed  route  varies  inversely  as  the  speed;  if  the 
speed  be  doubled,  the  train  will  traverse  its  route  in  one-half 
the  number  of  hours. 

351.  One  variable  number  is  said  to  vary  as  two  others  jointly 
when  it  varies  directly  as  their  product. 

Thus,  the  number  of  dollars  received  by  a  workman  in  a 
certain  number  of  days  varies  jointly  as  the  number  which  he 
receives  in  one  day,  and  the  number  of  days  during  which  he 
works. 

352.  One  variable  number  is  said  to  vary  directly  as  a  sec- 
ond and  inversely  as  a  third,  when  it  varies  jointly  as  the 
second  and  the  reciprocal  of  the  third. 


322  ALGEBRA 

Thus,  the  attraction  of  a  body  varies  directly  as  the  amount 
of  matter,  and  inversely  as  the  square  of  the  distance. 

353.  If  xtjzy,  then  x  equals  y  multiplied  by  a  constant  number. 
Let  x^  and  y'  denote  a  fixed  pair  of  corresponding  values  of 

X  and  y,  and  x  and  y  any  other  pair. 

By  the  definition  of  §  348,  -=  -,\   or,  a;  =  -7 y. 

J  '   y       y"         '  y'^ 

x' 
Denoting  the  constant  ratio  — ,  by  m,  we  have 

x  =  my. 

354.  It  follows  from  §§  350,  351,  352,  and  353  that : 

1.  If  X  varies  inversely  as  y^  x  =  — 

2.  Ifx  varies  jointly  as  y  and  z,  x  =  myz. 

my 

3.  Ifx  varies  directly  as  y  and  iiiversely  as  z,  x  =  — • 

355.  If  xccy,  and  y^z,  then  xocz. 

By  §  353,  ii  xccy,  x  =  my.  (1) 

And  if  2/  QC  2,  y  =  nz. 

Substituting  in  (1),  x  =  mnz. 

Whence,  by  §  353,  xazz. 

356.  If  xccy  when  z  is  constant,  and  xgcz  when  y  is  constant, 
then  xccyz  when  both  y  and  z  vary. 

Let  y'  and  z'  be  the  values  of  y  and  z,  respectively,  when  x 
has  the  value  x'. 

Let  y  be  changed  from  y'  to  y",  z  remaining  constantly  equal 
to  z',  and  let  x  be  changed  in  consequence  from  x'  to  X, 

Then  by  §348,  J=-^.  (1) 

Now  let  z  be  changed  from  z'  to  z",  y  remaining  constantly 
equal  to  ?/'',  and  let  x  be  changed  in  consequence  from  X  to  x". 


VARIATION  323 

Then,  |=^^.  (2) 

Multiplying  (1)  by  (2),  £;  =  lg.  (3) 

Now  if  both  changes  are  made,  that  is,  y  from  ?/'  to  2/"  and 
z  from  z'  to  z'\  x  is  changed  from  x'  to  ic",  and  yz  is  changed 
from  y'z'  to  2/"2!". 

Then  by  (3),  the  ratio  of  any  two  values  of  x  equals  the 
ratio  of  the  corresponding  values  of  yz ;  and,  by  §  348,  x  oc  yz. 

The  following  is  an  illustration  of  the  above  theorem  : 
It  is  known,  by  Geometry,  that  the  area  of  a  triangle  varies  as  the  base 
when  the  altitude  is  constant,  and   as   the   altitude   when   the  base  is 
constant ;  hence,  when  both  base  and  altitude  vary,  the  area  varies  as 
their  product. 

357.   Problems. 

Problems  in  variation  are  readily  solved  by  converting  the 
variation  into  an  equation  by  aid  of  §  §  353  or  354. 

1.  If  cc  varies  inversely  as  y,  and  equals  9  when  2/ =  8,  find 
the  value  of  x  when  y  =  l^. 

If  X  varies  inversely  as  y,  x  =  —  (§  354). 

Putting  cc  =  9  and  y  =  8,  9  =  - ,  or  m  =  72. 

8 

Then,  a;  =  ^ ;   and,  if  y  =  18,  a:  =  ^  =  4. 
y  18 

2.  Given  that  the  area  of  a  triangle  varies  jointly  as  its  base 
and  altitude,  what  will  be  the  base  of  a  triangle  whose  altitude 
is  12,  equivalent  to  the  sum  of  two  triangles  whose  bases  are 
10  and  6,  and  altitudes  3  and  9,  respectively  ? 

Let  B^  H,  and  A  denote  the  base,  altitude,  and  area,  respectively,  of 
any  triangle,  and  B'  the  base  of  the  required  triangle. 

Since  A  varies  jointly  as  B  and  H,  A  =  mBH  (§  354). 

Therefore,  the  area  of  the  first  triangle  is  m  x  10  x  3,  or  30  m,  and  the 
area  of  the  second  is  m  x  6  x  9,  or  54  m. 

Then,  the  area  of  the  required  triangle  is  30  w  -f-  54  m,  or  84  m. 


324  ALGEBRA 

But,  the  area  of  the  required  triangle  is  also  m  y.  B'  x  12. 
Therefore,  12  mB'  =  S4m,  or  B'  =7, 

EXERCISE  152 

1.  li  y  ccx,  and  x  equals  6  when  y  equals  54,  what  is  the 
value  of  y  when  x  equals  8  ? 

2.  If  X  varies  inversely  as  y,  and  equals  f  when  y  =  %,  what 
is  the  value  of  y  when  x  —  ^? 

3.  li  ycc  z^,  and  equals  40  when  z  =  10,  what  is  the  value  of 
y  in  terms  of  z^  ? 

4.  If  z  varies  jointly  as  x  and  y,  and  equals  f  when  a;  =  | 
and  2/  =  |>  what  is  the  value  of  z  when  x=  ^  and  y  =  ^? 

5.  If  X  varies  directly  as  y  and  inversely  as  z,  and  equals  -^^ 
when  y  —  21  and  2!  =  64,  what  is  the  value  of  x  when  2/  =  9  and 
2  =  32? 

6.  If  x^  oc  2/^,  and  a;  =  4  when  2/  =  4,  what  is  the  value  of  y 
when  a;  =  |-  ? 

7.  If  5  a;  +  8 oc 6 2/— 1?  and  a;=6  when  y=—3,  what  is  the 
value  of  X  when  y  =  7? 

8.  The  surface  of  a  cube  varies  as  the  square  of  its  edge.  If 
the  surface  of  a  cube  whose  edge  is  |-  feet  is  -^-^-  square  feet,  what 
will  be  the  edge  of  a  cube  whose  surface  is  ^^  square  feet  ? 

9.  If  5  men  in  6  days  earn  $  57,  how  many  days  will  it  take 
4  men  to  earn  $  76 ;  it  being  given  that  the  amount  earned 
varies  jointly  as  the  number  of  men,  and  the  number  of  days 
durjng  which  they  work. 

10.  The  volume  of  a  sphere  varies  jointly  as  its  diameter 
and  surface.  If  the  volume  of  a  sphere  whose  diameter  is  a, 
and  surface  b,  is  c,  what  is  the  diameter  of  a  sphere  whose  sur- 
face is  p  and  volume  q  ? 

11.  The  distance  fallen  by  a  body  from  rest  varies  as  the 
square  of  the  time  during  which  it  falls.  If  it  falls  579  feet 
in  6  seconds,  how  long  will  it  take  to  fall  402 J^  feet  ? 


VARIATION  325 

12.  A  circular  plate  of  lead,  17  inches  in  diameter,  is  melted 
and  formed  into  three  circular  plates  of  the  same  thickness. 
If  the  diameters  of  two  of  the  plates  are  8  and  9  inches  respec- 
tively, find  the  diameter  of  the  other;  it  being  given  that  the 
area  of  a  circle  varies  as  the  square  of  its  diameter. 

13.  If  y  equals  the  sum  of  two  numbers  which  vary  directly 
as  x^  and  inversely  as  x,  respectively,  and  2/  equals  —53  when 
X  equals  —  3,  and  ^-^-  when  x  equals  2,  what  is  the  value  of  y 
when  X  equals  ^^  ? 

14.  If  X  equals  the  sum  of  two  numbers,  one  of  which  varies 
directly  as  2/^  and  the  other  inversely  as  z^,  and  a?  =  45  when 
y  =  l  and  z  =  l,  and  ic  =  40  when  y  =  2  and  2;  =  3,  find  the  value 
of  y  when  ic  =  37  and  z—1. 

15.  If  y  equals  the  sum  of  three  numbers,  the  first  of  which 
is  constant,  and  the  second  and  third  vary  as  a?^  and  a?j  respec- 
tively, and  ?/  =  —  50  when  « =  2,  30  when  x  =  —  2,  and  110 
when  a;  =  —  3,  find  the  expression  for  y  in  terms  of  x. 

16.  The  volume  of  a  circular  coin  varies  jointly  as  its  thick- 
ness and  the  square  of  the  radius  of  its  face.  Two  coins  whose 
thicknesses  are  5  and  7  units,  and  radii  of  faces  60  and  30 
units,  respectively,  are  melted  and  formed  into  100  coins,  each 
3  units  thick.     Find  the  radius  of  the  face  of  the  new  coin. 

17.  The  weight  of  a  spherical  shell,  2  inches  thick,  is  ^f  of 
its  weight  if  solid.  Find  its  diameter,  it  being  given  that  the 
volume  of  a  sphere  varies  as  the  cube  of  its  diameter. 

PROBLEMS  IN  PHYSICS 

1.  When  the  force  which  stretches  a  spring,  a  straight  wire, 
or  any  elastic  body  is  varied,  it  is  found  that  the  displacement 
produced  in  the  body  is  always  directly  proportional  to  the 
force  which  acts  upon  it;  i.e.,  if  dj  and  dg  represent  any  two 
displacements,  and  j^  and  /g  respectively  the  forces  which  pro- 
duce them,  then  the  algebraic  statement  of  the  above  law  is 

-■=^-  (1) 


326  ALGEBRA 

If  a  force  of  2  pounds  stretches  a  given  wire  .01  inch,  how 
much  will  a  force  of  20  pounds  stretch  the  same  w^ire  ? 

2.  If  the  same  force  is  applied  to  two  wires  of  the  same 
length  and  material,  but  of  different  diameters,  D^  and  D2,  then 
the  displacements  di  and  d^  are  found  to  be  inversely  propor- 
tional to  the  squares  of  the  diameters,  i.e.. 

If  a  weight  of  100  kilograms  stretches  a  wire  .5  millimeter 
in  diameter  through  1  millimeter,  how  much  elongation  will 
the  same  weight  produce  in  a  wire  1.5  millimeters  in  diameter  ? 

3.  If  the  same  force  is  applied  to  two  wires  of  the  same 
diameter  and  material,  but  of  different  lengths,  l^  and  I2,  then  it 
is  found  that  ,      7 

f  =  f-  (3) 

Erom  (1),  (2),  and  (3)  and  §  356,  it  follows  that  when  lengths, 
diameters,  and  forces  are  all  different, 

^  =  ^  X  -  X  ^^'  (4) 

^2        J2         ^2         -^1 

If  a  force  of  1  pound  will  stretch  an  iron  wire  which  is 
200  centimeters  long  and  .5  millimeter  in  diameter  through 
1  millimeter,  what  force  is  required  to  stretch  an  iron  wire 
150  centimeters  long  and  1.25  millimeters  in  diameter  through 
.5  millimeter  ? 

4.  When  the  temperature  of  a  gas  is  constant,  its  volume 
is  found  to  be  inversely  proportional  to  the  pressure  to  which 
the  gas  is  subjected,  i.e.,  algebraically  stated, 

i^  =  S.  (5) 

At  the  bottom  of  a  lake  30  meters  deep,  where  the  pressure 
is  4000  grams  per  square  centimeter,  a  bubble  of  air  has  a  vol- 
ume of  1  cubic  centimeter  as  it  escapes  from  a  diver's  suit.  To 
what  volume  will  it  have  expanded  when  it  reaches  the  surface 
where  the  atmospheric  pressure  is  about  1000  grams  per  square 
centimeter  ? 


VARIATION  327 

5.  The  electrical  resistance  of  a  wire  varies  directly  as  its 
length  and  inversely  as  its  area.  If  a  copper  wire  1  centimeter 
in  diameter  has  a  resistance  of  1  unit  per  mile,  how  many  units 
of  resistance  will  a  copper  wire  have  which  is  500  feet  long 
and  3  millimeters  in  diameter? 

6.  The  illumination  from  a  source  of  light  varies  inversely 
as  the  square  of  the  distance  from  the  source.  A  book  which 
is  now  10  inches  from  the  source  is  moved  15  inches  farther 
away.     How  much  will  the  light  received  be  reduced  ? 

7.  The  period  of  vibration  of  a  pendulum  is  found  to  vary 
directly  as  the  square  root  of  its  length.  If  a  pendulum  1  meter 
long  ticks  seconds,  what  will  be  the  period  of  vibration  of  a 
pendulum  30  centimeters  long  ? 

8.  The  force  with  which  the  earth  pulls  on  any  body  out- 
side of  its  surface  is  found  to  vary  inversely  as  the  square  of 
the  distance  from  its  center.  If  the  surface  of  the  earth  is 
4000  miles  from  the  center,  what  would  a  pound  weight  weigh 
15,000  miles  from  the  earth  ? 

9.  The  number  of  vibrations  made  per  second  by  a  guitar 
string  of  given  diameter  and  material  is  inversely  proportional 
to  its  length  and  directly  proportional  to  the  square  root  of 
the  force  with  which  it  is  stretched.  If  a  string  3  feet  long, 
stretched  with  a  force  of  20  pounds,  vibrates  400  times  per 
second,  find  the  number  of  vibrations  made  by  a  string  1  foot 
long,  stretched  by  a  force  of  40  pounds. 

GRAPHS   IN  PHYSICS 

1.    Graphical  representation  of  a  direct  proportion. 

When  a  man  is  running  at  a  constant  speed,  the  distance 
which  he  travels  in  a  given  time  is  directly  proportional  to 
his    speed.     The    algebraic    expression   of    this    relation    is 

^  =  h^  ov  d  =  ms.     (See  §  353.) 

^2         ^2 


328 


ALGEBRA 


Now,  if  we  plot  successive  values  of  the  distance,  d,  which 
correspond  to  various  speeds,  s,  in  precisely  the  same  manner 
in  which  we  plotted  successive 
values  of  x  and  y  in  %  181,  we 
obtain  as  the  graphical  picture  of 
the  relation  between  s  and  d  a 
straight  line  passing  through  the 
origin.     (See  Fig.  1.) 

This  is  the  graph  of  any  direct 
proportion. 


Fig.  1. 


2.  Graphical  representation  of  an  inverse  proportion. 

The  volume  which  a  given  body  of  gas  occupies  when  the 
pressure  to  which  it  is  subjected  varies  has  been  found  to  be 
inversely  proportional  to  the  pressure  under  which  the  gas 
stands;  we  have  seen  that  the    algebraic  statement   of  this 

relation  is  — i  =  -~- 

If  we  plot  successive  values  of  V  and  P  in  the  manner  indi- 
cated in  §  181,  we  obtain  a  graph  of  the  form  shown  in  Fig.  2. 

This  is  the  graphical  representation 
of  any  inverse  j^roportion;  the  curve 
is  called  an  equilateral  hyperbola. 

3.  The  path  traversed  by  a  falling 
body  projected  horizontally. 

When  a  body  is  thrown  horizontally 
from  the  top  of  a  tower,  if  it  were  not 
for  gravity,  it  would  move  on  in  a 
horizontal  direction  indefinitely,  trav- 
ersing'  exactly  the  same  distance  in 
each  succeeding  second. 

Hence,  if  V  represents  the  velocity 
of  projection,  the  horizontal  distance, 
H,  which  it  would  traverse  in  any 
number  of  seconds,  t,  would  be  given 
by  the  equation  H=  Vt. 


or  V=- — . 


V=1, 


V=2, 


V=  3, 


P=' 


3* 


F=  4,       F= 


r= 


F= 


F= 


P 

5,  P- 

6,  P- 

T,  P: 

8,  P- 


VARIATION 


329 


On  account  of  gravity,  however,  the  body  is  pulled  down- 
ward, and  traverses  in  this  direction  in  any  number  of  seconds 
a  distance  which  is  given  by  the  equation  S  =  ^  gt^. 

To  find  the  actual  path  taken  by  the  body,  we  have  only  to 
plot  successive  values  of  H  and  S,  in  the  manner  in  which  we 
plotted  the  successive  values  of  x  and  y,  in  §  181. 

Thus,  at  the  end  of  1   second  the  vertical  distance  S^  is 

given   hj  Si  =  -g  xl^  =  -g;   at  the  end  of  2  seconds  we  have 


g ;  at  the  end  of  3  seconds,  S^  =  -g  y.2?  =-  g\ 


etc. 


1  -j  /» 

at  the  end  of  4  seconds,  /S'4  =  -^x4^  =  —  ^ 

On  the  other  hand,  at  the  end 
of  1  second  we  have  ^i=F;  at 
the  end  of  2  seconds,  ^2  =  2  F;  at 
the  end  of  3  seconds,  ^3  =  3  F; 
at  the  end  of  4  seconds,  ^4  =  4  F 

If,  now,  we  plot  these  successive 
values  of  H  and  >S',  we  obtain  the 
graph  shown  in  Fig.  3. 

This  is  the  path  of  the  body;  it 
is  a  parabola.     (§  314,  Ex.  2.) 


4.  Orapli  of  relation  between  the 
temperature  and  pressure  existing 
within  an  air-tight  boiler  containing  only  water  and  water  vapor. 

One  use  of  graphs  in  physics  is  to  express  a  relation  which 
is  found  by  experiment  to  exist  between  two  quantities,  which 
cannot  be  represented  by  any  simple  algebraic  equation. 

For  example,  when  the  temperature  of  an  air-tight  boiler 
which  contains  only  water  and  water  vapor  is  raised,  the  pres- 
sure within  the  boiler  increases  also ;  thus  we  find  by  direct 
experiment  that  when  the  temperature  of  the  boiler  is  0°  centi- 
grade, the  pressure  which  the  vapor  exerts  will  support  a 
column  of  mercury  4.6  millimeters  high. 


Fig.  3. 


330 


ALGEBRA 


When  the  temperature  is  raised  to  10°,  the  mercury  column 
rises  to  9.1  millimeters;  at  30°  the  column  is  31.5  milli- 
meters long,  etc. 

To  obtain  a  simple  and  compact  picture  of  the  relation 
between  temperature  and  pressure,  we  plot  a  succession  of 
temperatures,  e.g.  0°,  10°,  20°,  30°,  40°,  50°,  60°,  70°,  80°,  90°, 
100°,  in  the  manner  in  which 
we  plotted  successive  values  of 
a;  in  §  181,  and  then  plot  the 
corresponding  values  of  pres- 
sure obtained  by  experiment  in 
the  manner  in  which  w^e  plotted 
the  ^'s  in  §  181 ;  we  obtain  the 
graph  shown  in  Fig.  4. 

From  this  graph  we  can  find 
at  once  the  pressure  which  will 
exist  within  the  boiler  at  any 
temperature. 

For  example,  if  we  wish  to  know  the  pressure  at  75°  cen- 
tigrade, we  observe  where  the  vertical  line  which  passes 
through  75°  cuts  the  curve  and  then  run  a  horizontal  line 
from  this  point  to  the  point  of  intersection  with  the  line  OP. 

This  point  is  found  to  be  at  288 ;  hence  the  pressure  within 
the  boiler  at  75°  centigrade  is  288  millimeters. 


70    80    90  100 


PROGRESSIONS  331 


^^  XXVII.     PROGRESSIONS 

ARITHMETIC  PROGRESSION 

358.  An  Arithmetic  Progression  is  a  series  of  terms  in  which 
each  term,  after  the  first,  is  obtained  by  adding  to  the  preced- 
ing term  a  constant  number  called  the  Common  Difference. 

Thus,  1,  3,  5,  7,  9,  11,  •••  is  an  arithmetic  progression  in 
which  the  common  difference  is  2^ 

Again,  12,  9,  6,  3,  0,  —3,  •••  is  an  arithmetic  progression  in 
which  the  common  difference  is  —  3. 

An  Arithmetic  Progression  is  also  called  an  Arithmetic  Senes. 

359.  Given  the  first  term,  a,  the  common  difference,  d,  and  the 
number  of  terms,  n,  to  find  the  last  term,  I. 

The  progression  is  a,  a  -h  c?,  a  +  2  d,  a  +  3  cZ,  •••. 
We  observe  that  the  coefficient  of  d  in  any  term  is  less  by  1 
than  the  number  of  the  term. 

Then,  in  the  nth  term  the  coefficient  of  d  will  be  n  —  1. 

That  is,  l  =  a-\-{n-l)d.  (I) 

360.  Given  the  first  term,  a,  the  last  term,  I,  and  the  number  of 
terms,  n,  to  find  the  sum  of  the  terms,  S. 

8  =  a  +  {a  +  d)  +  {a+2d)-\-"'^{l-d)-{-L 
Writing  the  terms  in  reverse  order, 

S  =  l  +  {l-d)  +  il-2d)  +  '.'  +  {a  +  d)+a. 
Adding  these  equations  term  by  term, 

2^=(a  +  0  +  (a  +  0  +  («  +  0  +  -  +  («  +  0  +  («  +  0- 
Therefore,    2S^n(a  +  l),  and  S  =  '^(a  +  l),  (II) 

361.  Substituting  in  (II)  the  value  of  I  from  (I),  we  have 

5  =  |[2a  +  («-l)d]. 


332  ALGEBRA 

362.  Ex.  Ill  the  progression  8,  5,  2,  —1,  —4,  •..,  to  27 
terms,  find  the  last  term  and  the  sum. 

Here,  a  =  8,  (^  =  5  -  8  =  -  3,  w  =  27. 

Substitute  in  (I),       z  =  8  +  (27  -  1) ( ^  3)  =  8  -  78  =  -  70. 
Substitute  in  (II),    ,S'  =  —  (8  -  70)  =  27  (-  31)  =  -  837. 

The  common  difference  may  be  found  by  subtracting  the  first  term 
from  tlie  second,  or  any  term  from  the  next  following  term. 

EXERCISE  153 

In  each  of  the  following,  find  the  last  term  and  the  sum : 
1.   4,  9,  14,  .-.  to  14  terms.  2.    9,  2,  -5,  •••  to  16  terms. 

3.  -  51,  -  45,  -  39,  ...  to  15  terms. 

4.  -I,  --V-,  -3,  ...  to  13  terms. 

5.  hh  -h  '"  "to  18  terms.  6.  |,  JJ,  fi  ...  to  17  terms. 
*  7.    --V-,  -\%y  -1,  —  to  27  terms. 

8-    -TO?  -i  -Ii  •••  to  52  terms. 
9.    3a +  4&,  8a +  26,  13a,  ...to  10  terms. 

10.    ^Zl^I,   ^-,  ^^2/,  ...to  9  terms. 
3      '   6'    3  ' 

363.  The  first  term,  common  difference,  number  of  terms,  last 
term,  and  sum  of  the  terms,  are  called  the  elements  of  the 
progression. 

If  any  three  of  the  five  elements  of  an  arithmetic  progres- 
sion are  given,  the  other  two  may  be  found  by  substituting  the 
known  values  in  the  fundamental  formulae  (I)  and  (II),  and 
solving  the  resulting  equations. 

1.    Given  a  =  —  f ,  n  =  20,  /S  =  —  f ;  find  d  and  I. 
Substituting  the  given  values  in  (II), 


_§^10f-^4-?V  or-l  =  -^+Z;  then,  I  =  ^- -- 
3  V     3       /  6         3        '  '         3     6 


3 

6      2 


PROGRESSIONS  333 

3         5 

Substituting  tlie  values  of  a,  w,  and  I  in  (I),  -  =  — -+  19  ci. 

2         3 

Whence,  i9dJ  =  §  +  ^  =  ^,  and  <?  =  1. 

'  2     3      6  6 

2.    Given  d  =  -3,  Z  =  -39,  /S^  =  -264;  find  a  and  n. 
Substituting  in  (I),    -  39  =  a  +  (n  -  1)(-  3),  or  a  =  3  w  -  42.      (1) 
Substituting  tlie  values  of  Z,  /S",  and  a  in  (II) , 
-  264  =  -  (3  n  -  42  -  39),  or  -  528  =  3  n^  -  81  n,  or  ii^  _  27  w  +  176  =  0. 

2i 

„.,                           27  ±  V729  -  704      27  ±  5      -,^^11 
Whence,         n  =  — — =  — ='^—  =  16  or  11. 

'  2  2 

Substituting  in  (1),  a  =  48  -  42  or  33  -  42  =  6  or  -  9. 

The  solution  is  a  =  6,  n  =  16  ;  or,  a  =  —  9,  n  =  11. 

The  significance  of  the  two  answ^ers  is  as  follows  : 

If  a  =  6  and  n  =  16,  the  progression  is  6,  3,  0,   —  3,  — 6,   —  9,   —  12, 
15,   _18,   -21,   -24,   -27,   -30,   -33,   -36,   -39. 

If  a  =  —  9  and  w  =  11,  the  progression  is 

-9,   -12,   -15,   -18,   -21,   -24,   -27,   -30,   -33,   -36,   -39. 

In  each  of  these  the  sum  is  —  264. 

3:   Given  a=^,  d  =  —  ^^y  ^  =  —  | ;  find  I  and  n. 

Substitutmg  in  (I),  Z  =  1  +  (71  -  1)  (  -  ^)  =  ^^  (1) 

Substituting  the  values  of  a,  S,  and  I  in  (II), 


-i(h^)'  ''  -'  =  "('-12^)'  ''  -^-9--36  =  0. 


Whence,         ^  ^  9  j:  VsTTU? ^ 9^15  ^  1,  ^,  _3^ 

2  2 

The  value  w  =  —  3  must  be  rejected,  for  the  number  of  terms  in  a 
progression  must  be  a  positive  integer. 

5-12  7 


Substituting  w  =  12  in  (1),  1  = 


12  12 


A  negative  or  fractional  value  of  n  must  be  rejected,  together  with  all 
other  values  dependent  on  it. 


334  ALGEBRA 

EXERCISE  154 

1.  Given  d  =  S,  ^  =  115,  n  =  15;  find  a  and  S. 

2.  Given  d  =  -6,  71  =  14,  aS  =  -616;  find  a  and  L 

3.  Given  a  =  -69,  n  =  16,  Z  =  36;  find  d  and  S. 

4.  Given  a  =8,  ?i  =  25,  >S  =  -2500;  find  d  and  ?. 

5.  Given  a  =  f,  l  =  -\\  /S'  =  -78;  find  d  and  71. 

6.  Given  Z  =  H^,  n  =  13,  AS  =  -2^f^;  find  a  and  d 

7.  Given  a  =  — f,  d  =  — ^,  /S'  =  — ^^;  find  n  and  I 

8.  Given  a  =  — f,  1  =  ^,  <^  =  |;  find  n  and  >S^. 

9.  Given  d  =  —  ^,  n  =  55,  S  —  —  165;  find  a  and  Z. 

10.  Given  ?  =  ^V-,  n  =  24,  iS  =  241;  find  a  and  c?. 

11.  Given  Z  =  -9/,  d  =  -|,  .^  =  ^f^;  find  a  and  n. 

12.  Given  a  =  -^,  l  =  -U,  S  =  -^^;  find  c?  and  n. 

13.  Given  a  — —  ^2,  n  =  21,  /S"  =  || ;  find  c?  and  I. 

14.  Given  ?  =  |f,  d  =  ^,  8  =  -^^^;  find  a  and  n. 

15.  Given  a  =  -%S  d  =  i    aS  =  -^-;  fiii^  n  and  /. 

364.   From  (I)  and  (II),  general  formulce  for  the  solution  of 
examples  like  the  above  may  be  readily  derived. 

Ex.     Given  a,  d,  and  ^iS ;  derive  the  formula  for  n. 

By  §  361,  2  ^S'  =  ri[2  a  +  (n  -  l)c?],  or  dn"^  -\-  {2  a  -  d)n  =  2  S. 
This  is  a  quadratic  in  n,  and  may  be  solved  by  the  method  of  §  288  ; 
multiplying  by  4  d,  and  adding  (2  a  —  dy  to  both  members, 

4  d2n2  +  4  d(2  a  -  (^)?z  +  (2  a  -  (?)2  =  8  d^  +  (2  a  -  d)2. 


Extracting  square  roots,     2dn  +  2a  —  <?=  ±  V8  d/S"  +  (2  a  —  d)^. 


Whence,  n  =  d -2a  ±  VSdS -^  {2a  -  d)\. 

2d 


EXERCISE    155 

1.   Given  a,  I,  and  n ;  derive  the  formula  for  d. 


PROGRESSIONS  335 

2.  Given  a,  n,  and  S;  derive  the  formulae  for  d  and  I. 

3.  Given  d,  n,  and  S ;  derive  the  f ormulie  for  a  and  I.    , 

4.  Given  a,  dj  and  / ;  derive  the  formulae  for  n  and  S. 

5.  Given  d,  I,  and  n ;  derive  the  formulae  for  a  and  JS. 

6.  Given  I,  n,  and  ^ ;  derive  the  formulae  for  a  and  d. 

7.  Given  a,  d,  and  /iS ;  derive  the  formula  for  I. 

8.  Given  a,  Z,  and  ^S ;  derive  the  formulae  for  d  and  n. 

9.  Given  d,  I,  and  aS;  derive  the  formulae  for  a  and  n. 

365.  Arithmetic  Means. 

We  define  inserting  m  arithmetic  means  between  two  given 
numbers,  a  and  b,  as  finding  an  arithmetic  progression  of  m  +  2 
terms,  whose  first  and  last  terms  are  a  and  b. 

Ex.     Insert  5  arithmetic  means  between  3  and  —  5. 

We  find  an  arithmetic  progression  of  7  terms,  in  which  a  =  3,  and 
l  =  ~5;  substituting  n  =  7,  a  =  3,  and  Z  =  —  5  in  (I), 

-5  =  3  +  6^,  or  d  =  -^- 

o 

The  progression  is  3,  |,    ^,    -1,    -I    -^,   -5. 
o      o  o  o 

366.  Let  X  denote  the  arithmetic  mean  between  a  and  b. 
Then,  x  —  a  =  b  —  x,  or  2x  =  a-\-b. 

Whence,  x  =  ^^±-5 . 

'  •  2 

That  is,  the  arithmetic  mean  between  two  numbers  equals  one- 
half  their  sum. 

EXERCISE    156 

1.  Insert  7  arithmetic  means  between  4  and  10. 

2.  Insert  6  arithmetic  means  between  —  |  and  —  Y-- 

3.  Insert  9  arithmetic  means  between  —  ^  and  6. 


336  ALGEBRA 

4.  Insert  8  arithmetic  means  between  —  3  and  —  ^. 

5.  Insert  5  arithmetic  means  between  f  and  —  ^. 

6.  How  many  arithmetic  means  are  inserted  between  —  | 
and  j-^,  when  the  sum  of  the  second  and  last  is  f? 

7.  If  m  arithmetic  means  are  inserted  between  a  and  b,  find 
the  first  two. 

Find  the  arithmetic  mean  between : 

8.  V-  a^^  -  ¥•  9.    (3  m  +  nf  and  (m  -  3  nf. 

367.   Problems. 

1.    The  sixth  term  of  an  arithmetic  progression  is  f ,  and  the 
fifteenth  term  is  ^-.     Find  the  first  term. 

By  §  359,  the  sixth  term  is  a  +  5  d,  and  the  fifteenth  term  a-\-  lid. 


o 
\a  +  Ud  =  f 

(1) 

Then  by  the  conditions, 

(2) 

Subtracting  (1)  from  (2), 

9d  =  -;  whence,  d  =  -. 
2'                           2 

Substituting  in  (1), 

^"^i"i'^^^^°^'^"~i' 

2.  Find  four  numbers  in  arithmetic  progression  such  that 
the  product  of  the  first  and  fourth  shall  be  45,  and  the  product 
of  the  second  and  third  77. 

Let  the  numbers  be  x  —  Sy,  x  —  y,  x  +  y,  and  x  +  Sy. 

-  9  ?/2  =  45. 


Then  by  the  conditions,    ,    ,  „      „„ 

Solving  these  equations,  x  =  9,  y  =±2  ;  or,  a;  =  —  9,  ?/  =  ±  2  (§  308). 
..Then  the  numbers  are  3,  7,  11,  15;   or,   —3,   —7,   —11,   —15. 

In  problems  like  the  above,  it  is  convenient  to  represent  the  unknown 
numbers  by  symmetrical  expressions. 

Thus,  if  five  numbers  had  been  required,  we  should  have  represented 
them  by  ic  —  2  y,  x  —  y,  x,  x  -^  y,  and  x  -\-  2y. 


PROGRESSIONS  337 

EXERCISE  157 

1.  The  fifth  term  of  an  arithmetic  progression  is  ^,  and 
the  thirteenth  term  |.     Find  the  twenty-second  term. 

2.  Find  the  sum  of  all  the  odd  integers,  beginning  with  1 
and  ending  with  999. 

3.  How  many  positive  integers  of  three  digits  are  multiples 
of  7  ?     What  is  their  sum  ? 

4.  The  first  term  of  an  arithmetic  progression  is  1,  and  the 
sum  of  the  sixth  and  tenth  terms  is  37.  Find  the  second  and 
third  terms. 

5.  The  first  term  of  an  arithmetic  progression  of  11  terms 
is  |,  and  the  seventh  term  —  3.     Find  the  sum  of  the  terms. 

6.  In  an  arithmetic  progression,  the  sum  of  the  first  and 
last  terms  is  two-ninths  the  sum  of  all  the  terms.  Find  the 
number  of  terms. 

7.  The  seventh  term  of  an  arithmetic  progression  is  —  37, 
and  the  sum  of  the  first  17  terms  —  799.  Find  the  sum  of  the 
first  13  terms. 

8.  Find  five  numbers  in  arithmetic  progression  such  that 
the  sum  of  the  first,  fourth,  and  fifth  is  14,  and  the  quotient  of 
the  second  by  the  fourth  —  ^. 

9.  How  many  arithmetic  means  are  inserted  between  —  | 
and  I,  when  their  sum  is  -y-  ? 

10.  If  the  constant  difference  of  an  arithmetic  progression 
equals  twice  the  first  term,  the  quotient  of  the  sum  of  the  terms 
by  the  first  term  equals  the  square  of  the  number  of  terms. 

11.  The  sum  of  the  first  10  terms  of  an  arithmetic  progression 
is  to  the  sum  of  the  first  5  terms  as  13  to  4.  Find  the  ratio  of 
the  first  term  to  the  common  difference. 

12.  Find  four  numbers  in  arithmetic  progression  such  that 
the  sum  of  the  first  and  second  shall  be  —1,  and  the  product 
of  the  second  and  fourth  24. 


338  ALGEBRA 

13.  The  last  term  of  an  arithmetic  progression  of  10  terms 
is  29.  The  sum  of  the  odd-numbered  terms  is  to  the 
sum  of  the  even-numbered  terms  as  14  is  to  17.  Find  the  first 
term  and  the  common  differ'ence. 

14.  The  sum  of  five  numbers  in  arithmetic  progression  is  25, 
and  the  sum  of  their  squares  is  135.     Find  the  numbers. 

15.  A  man  travels  ^^  miles.  He  travels  10  miles  the  first 
day,  and  increases  his  speed  one-half  mile  in  each  succeeding 
day.     How  many  days  does  the  journey  require  ? 

16.  Find  the  sum  of  the  terms  of  an  arithmetic  progression 
of  9  terms,  in  which  17  is  the  middle  term. 

17.  Find  three  numbers  in  arithmetic  progression,  such  that 
the  square  of  the  first  added  to  the  product  of  the  other  two 
gives  16,  and  the  square  of  the  second  added  to  the  product  of 
the  other  two  gives  14. 

18.  If  a  person  saves  $  120  each  year,  and  puts  this  sum  at 
simple  interest  at  3|%  at  the  end  of  each  year,  to  how  much 
will  his  property  amount  at  the  end  of  18  years  ? 

19.  A  traveller  sets  out  from  a  certain  place,  and  goes 
7  miles  the  first  hour,  7^  the  second  hour,  8  the  third  hour, 
and  so  on.  After  he  has  been  gone  5  hours,  another  sets  out, 
and  travels  16J  miles  an  hour.  How  many  hours  after  the 
first  starts  are  the  travellers  together  ? 

20.  There  are  12  equidistant  balls  in  a  straight  line.  A 
person  starts  from  a  position  in  line  with  the  balls,  and  beyond 
them,  his  distance  from  the  first  ball  being  the  same  as  the 
distance  between  the  balls,  and  picks  them  up  in  succession, 
returning  with  each  to  his  original  position.  He  finds  that  he 
has  walked  780  feet.     Find  the  distance  between  the  balls. 

GEOMETRIC  PROGRESSION 

368.  A  Geometric  Progression  is  a  series  of  terms  in  which 
each  term,  after  the  first,  is  obtained  by  multiplying  the 
preceding  term  by  a  constant  number  called  the  Ratio. 


PROGRESSIONS  339 

Thus,  2,  6, 18,  54, 162,  •••  is  a  geometric  progression  in  which 
the  ratio  is  3. 

9,  3,  1,  i,  -J-,  •••  is  a  geometric  progression  in  which  the  ratio 
isi. 

—  3,  6,  — 12,  24,  —  48,  •••  is  a  geometric  progression  in  which, 
the  ratio  is  —  2. 

A  Geometric  Progression  is  also  called  a  Geometric  Series. 

369.  Given  the  first  term,  a,  the  ratio,  r,  and  the  number  of 
terms,  n,  to  find  the  last  term,  I. 

The  progression  is  a,  ar,  ar^,  ai^,  •  •  • . 

We  observe  that  the  exponent  of  r  in  any  term  is  less  by  1 
than  the  number  of  the  term. 

Then,  in  the  wth  term  the  exponent  of  r  will  be  n  —  1. 

That  is,  I  =  ar"-\  (I) 

370.  Given  the  first  term,  a,  the  last  term,  I,  and  the  ratio,  r,  to 
find  the  sum  of  the  terms,  S. 

S  =  a  +  ar  +  ar'-  -\ h  ar""-^  +  ar"-^  +  ar~-^  (1) 

Multiplying  each  term  by  r, 

rS  =  ar-{-  ai^  +  ar^  +  •  •  •  +  ar""-^  +  ar""^  +  ar"".  (2) 

Subtracting  (1)  from  (2),  r8-S  =  ar^-a,  or  8  =  ^^^~^' 

But  by  (I),  §  369,  rl=^ar\ 

Therefore,  ^  =  'l:z^.  (II) 

r  —  1 

The  first  term,  ratio,  number  of  terms,  last  term,  and  sum  of  the  terms, 
are  called  the  elements  of  the  progression. 

371.  Examples. 

1.  In  the  progression  3,  J,  •!■,•••,  to  7  terms,  find  the  last 
term  and  the  sum. 

Here,  a  =  S,  r  =  -,  n  =  7. 


340  .  ALGEBRA 

Substituting  in  (I),        I  =  ^(^)'=^5  =  ^- 

lxA-_3      J__3  2186 

o  ^.  .-.  .•       •     /TT^        c      ^      243             729                  729       1093 
Substituting  in  (II),      S  =  — = ^  = ^  =  -^  - 

S~  ~3  ~3 

The  ratio  may  be  found  by  dividing  the  second  term  by  the  first,  or 
any  term  by  the  next  preceding  term. 

2.    In  the  progression  —  2,  6,  — 18,  •••,  to  8  terms,  find  the 
last  term  and  the  sum. 

Here,  a  =  —  2,  r  = =  —  3,  w  =  8  ;  therefore, 

;  =  _  2(-  3)7  =  -  2  X  (-  2187)  =  4374. 

,,      -  3  X  4374  -(-2)      -  13122  +  2     000^ 

^  =  —  _3-i =  ~-~:^ — =^^^^-    ■ 


EXERCISE  158 

Eind  the  last  term  and  the  sum  of  the  following : 

1.  1,  —  2,  4,  ...  to  10  terms.  6.  —  f,  i,  —  h  •••  to  7  terms. 

2.  -6, -9,-^-,  •••  to  7  terms.     7.  -4, -3, -|,  ••.  to5terms. 

3.  3,  — 15,  75,  ...  to  5  terms.         8.  —  |,  f,  —  -VS  •••  to  8  terms. 

4.  -5, -20, -80,  ...to  6  terms.     9.  2,  f ,  2%,  ...  to  6  terms. 

5.  ^,  i,  1,  ...  to  9  terms.  10.  f,  —J,  i,  ...  to  8  terms. 

372.  If  any  three  of  the  five  elements  of  a  geometric  pro- 
gression are  given,  the  other  two  may  be  found  by  substituting 
the  given  values  in  the  fundamental  formulae  (I)  and  (II),  and 
solving  the  resulting  equations. 

But  in  certain  cases  the  operation  involves  the  solution  of  an 
equation  of  a  degree  higher  than  the  second ;  and  in  others  the 
unknown  number  appears  as  an  exponent,  the  solution  of  which 
form  of  equation  can  usually  only  be  effected  by  the  aid  of 
logarithms  (§  437). 

In  all  such  cases  in  the  present  chapter,  the  equations  may 
be  solved  by  inspection. 


PROGRESSIONS  341 

1.  Given  a  =  —  2,  w  =  5,  Z  =  —  32^  find  r  and  S. 
Substituting  the  given  values  in  (I),  we  have 

—  32  =  —  2  r* ;  whence,  r*  =  16,  or  r  =  ±  2. 

Substituting  in  (II), 

If  r=     2,  ^  =  2(-^2)-(-2)^_g4_^2=:_62. 

z  —  1 

If  r  =  -2    ^^(-2)(-32)-(-2)^64  +  2^_22^ 

-2-1  -3 

The  solution  is  r  =  2,  /S^  =  -  62  ;   or,  r  =  -  2,  /S'  =  -  22. 

The  interpretation  of  the  two  answers  is  as  follows  : 

If  r=     2,  the  progression  is  —2,  —4,  —8,  —16,  —32,  whose  sum  is  —62. 

If  r=  —2,  the  progression  is  —2,     4,  —8,      16,  —32,  whose  sum  is  —22. 

2.  Given  a  =  3,  r  =  -i   8  =  ^^-^^,   find  yi  and  ?. 

-lz-3 
Substituting  in  (II),       ^  =  _^  =  ^. 

~3~ 

WhencP  7_L9-6560.  6560      q_       1 

Whence,  z  +  9 --^  ,  or,  Z  _ -^      9_     ^^^. 

Substituting  the  values  of  a,  r,  and  Z  in  (I) , 

__L=3f-lv-^or,  f-iv-^=-^. 

729        V     3/       '      '  V     3;  2187 

Whence,  by  inspection,  w  —  1  =  7,  or  n  =  S. 

From  (I)  and  (II)  general  formulae  may  be  derived  for  the  solution  of 
cases  like  the  above. 

If  the  given  elements  are  n,  I,  and  S,  equations  for  a  and  r  may  be 
found,  but  there  are  no  definite  formulae  for  their  values. 

The  same  is  the  case  when  the  given  elements  are  a,  w,  and  S. 

The  general  formulae  for  n  involve  logarithms ;  these  cases  are  discussed 
in  §  437. 

EXERCISE  159 

1.  Given  r  =  3,  n  =  %,l  =  2187 ;  find  a  and  S. 

2.  Given  r  =  —  4,  n  =  5,  S  =  —  410 ;  find  a  and  I. 


342  ALGEBRA 

3.  Givena  =  6,n  =  6,  Z  =  -f4;  find  rand  >S. 

4.  Given  a  =  3,  r=^,  1  =  y|^ ;  find  n  and  S. 

5.  Given  r  =  -2,  71  =  10, /S  =  -i-V>-^;  find  a  and  L 

6.  Given  a  =  f ,  n  =  7,  ?  =  f f ;  find  ?-  and  S. 

7.  Given  a  =  -i  ?  =  --2/^,/S'  =  -fff;  find  r  and  w. 

8.  Given  a  =  |,  r  =  |,  ,S  =  mf ;  find  I  and  n. 

9.  Given  I  =  -  768,  r  =  4,  .S  =  -  ^Q^^ ;  find  a  and  n. 

10.  Given  a  =  |,  Z  =  1458,  S=  ^-^ ;  find  r  and  w. 

11.  Given  a,  r,  and  /S;  derive  the  formula  for  I. 

12.  Given  a,  Z,  and  aS  ;  derive  the  formula  for  r. 

13.  Given  r,  I,  and  8;  derive  the  formula  for  a. 

14.  Given  r,  n,  and  I ;  derive  the  formulae  for  a  and  S. 

15.  Given  r,  n,  and  S\  derive  the  formulae  for  a  and  I. 

16.  Given  a,  n,  and  Z ;  derive  the  formulae  for  r  and  8. 

373.   Sum  of  a  Geometric  Progression  to  Infinity. 

The  limit  (§  318)  to  which  the  sum  of  the  terms  of  a  decreas- 
ing geometric  progression  approaches,  when  the  number  of 
terms  is  indefinitely  increased,  is  called  the  sum  of  the  series 
to  infinity. 

Formula  (II),  §  370,  may  be  written 

a  —  rl  ^ 


8  = 


1-r 


It  is  evident  that,  by  sufficiently  continuing  a  decreasing 
geometric  progression,  the  absolute  value  of  the  last  term  may 
be  made  less  than  any  assigned  number,  however  small. 

Hence,  when  the  number  of  terms  is  indefinitely  increased, 
Z,  and  therefore  rl,  approaches  the  limit  0. 

Then,  the  fraction  ^~^   approaches  the  limit  — — 
1— r  1—r 


PROGRESSIONS  343 

Therefore,  the  sum  of  a  decreasing  geometric  progression  to 
infinity  is  given  by  the  formula 

S  =  -^^'  (III) 

1  —  r 

Ex.     Find  the  sum  of  the  series  4,  —  |,  ^-j  -"to  infinity. 


Here,  a  =  4,  r  =  — 
3 


2 

4         12 


Substituting  in  (III),  S ^  -  - 

1+1      6 


EXERCISE  160 

Find  the  sum  to  infinity  of  the  following : 

1-   6,2,1,....  5.   hU,U,-- 

2.   12,-3,1,....  6.    -J^,^%., -^^,  .... 

.^111...  7_3_15  75       ... 

4       —25      25      _   50     ...  8       5      _    5         10       ... 

374.    To  j^nd  ^^e  value  of  a  repeating  decimal. 
This  is  a  case  of  finding  the  sum  of  a  decreasing  geometric 
series  to  infinity,  and  may  be  solved  by  formula  (III). 

Ex.     Find  the  value  of  .85151  .... 

We  have,  .85151  ...  =  .8  +  .051  +  .00051  +  .... 

The  terms  after  the  first  constitute  a  decreasing  geometric  progression, 
in  which  a  =  .051,  and  r  =  .01. 

.051        .051      51       17 


Substituting  in  (III),  S 


.01       .99      990     330 


8        17  281 

Then,  the  value  of  the  given  decimal  is 1 ,  or 

^  10     330        330 


EXERCISE  161 
Find  the  values  of  the  following : 

1.  .7272  ....  3.    .91777  ....  5.   .23135135  • 

2.  .629629  ....  4.    .75959  ....  6.   .587474  .... 


344  ALGEBRA 

375.  Geometric  Means. 

We  define  inserting  m  geometric  means  between  two  numbers, 
a  and  b,  as  finding  a  geometric  progression  of  m  +  2  terms, 
whose  first  and  last  terms  are  a  and  b. 

Ex.   Insert  5  geometric  means  between  2  and  ^|f . 

We  find  a  geometric  progression  of   7   terms,  in  which  a  =  2,  and 

128  128  ' 

I  =  — ;  substituting  w  =  7,  a  =  2,  and  I  =  — —  in  (I), 

729  729  ' 

1|  =  2^;  whence  r«  =  ^,  and.  =  ±| 

Theresultis2,  ±^,   8         16     32         64      128. 
'3     9         27     81'       243'   729 

376.  Let  X  denote  the  geometric  mean  between  a  and  b. 

Then,  -  =  -,  or  ic^  =  ab. 

a     X 

Whence,  x  =  Va&. 

That  is,  the  geometric  mean  between  two  numbers  is  equal  to 
the  square  root  of  their  product. 

EXERCISE  162 

1.  Insert  4  geometric  means  between  f  and  24. 

2.  Insert  5  geometric  means  between  —  3  and  —  2187. 

3.  Insert  4  geometric  means  between  -^  and  —  320. 

4.  Insert  6  geometric  means  between  —  f  and  ^^\. 

5.  Insert  7  geometric  means  between  —  48  and  —  -^^. 

6.  Insert  3  geometric  means  between  J^^-  and  ■^^. 

7.  If  m  geometric  means  are  inserted  between  a  and  b,  what 
are  the  last  two  means  ? 

Find  the  geometric  mean  between  : 


8.   -U  and  Y-  9.    ^^  and  ^^^. 

^8  xy-y^  xy 

10.   a2-4a  +  4  and  4a2  +  4a  +  l. 


PROGRESSIONS  345 

377.   Problem. 

Find  3  numbers  in  geometric  progression  such  that  their 

sum  shall  be  14,  and  the  sum  of  their  squares  84. 

Let  the  numbers  be  represented  by  a,  ar,  and  ar^. 

r       a  +  ar+ar2  =  14.  ^  (1) 

Then,  by  the  conditions,  < 

'    ^  '    I  a2  +  a-V2  +  aV  =  84.  (2) 

Divide  (2)  by  (1),  a  -  ar  +  ar2  =  6.  (3) 

Subtract  (3)  from  (1),  2  ar  =  8,  or  r  =  -•  (4) 

a 

Substituting  in  (1),  a  +  4 +  — =  14,  or  a2  -  10  a  +  16=0. 

a 


Solving  this  equation,  a  =  8  or  2. 

4  4 
-  or  - 
8        2     2 


4        4      1 
Substituting  in  (4),  r  =  -  or  -  =  -  or  2. 


Then,  the  numbers  are  2,  4,  and  8. 

EXERCISE  163 

1.  The  fifth  term  of  a  geometric  progression  is  |,  and  the 
eighth  term  —  -^^.     Find  the  third  term. 

2.  The  product  of  the  first  five  terms  of  a  geometric  pro- 
gression is  243.     Find  the  third  term. 

3.  Find  four  numbers  in  geometric  progression  such  that 
the  sum  of  the  first  and  fourth  shall  be  27,  and  of  the  second 
and  third  18. 

4.  Find  an  arithmetic  progression  whose  first  term  is  2,  and 
whose  first,  fourth,  and  tenth  terms  form  a  geometric  pro- 
gression. 

5.  The  third  term  of  a  geometric  progression  is  f ,  and  the 
seventh  term  -Y^-^'t  find  the  ninth  term. 

6.  The  sum  of  the  terms  of  a  geometric  progression  whose 
first  term  is  1,  ratio  3,  and  number  of  terms  4,  equals  the  sum 
of  the  terms  of  an  arithmetic  progression  whose  first  term  is 
4,  and  common  difference  4.  Find  the  number  of  terms  of  the 
arithmetic  progression. 


346  ALGEBRA 

7.  The  sum  of  the  first  four  terms  of  a  decreasing  geometric 
progression  is  to  the  sum  to  infinity  as  16  to  25.  Find  the 
ratio. 

8.  A  man  who  saved  every  year  four-thirds  as  much  as  in 
the  preceding  year,  had  in  four  years  saved  $  3500.  How- 
much  did  he  save  the  first  year  ? 

9.  The  difference  between  two  numbers  is  16,  and  their 
arithmetic  mean  exceeds  their  geometric  mean  by  2.  Find  the 
numbers. 

10.  Find  six  numbers  in  geometric  progression  such  that  the 
sum  of  the  first  and  fourth  shall  be  9,  and  of  the  third  and 
sixth  36. 

11.  The  digits  of  a  number  of  three  figures  are  in  geometric 
progression,  and  their  sum  is  7.  If  297  be  added  to  the  num- 
ber, the  digits  will  be  reversed.     Find  the  number. 

12.  There  are  three  numbers  in  geometric  progression  whose 
sum  is  ^.  If  the  first  be  multiplied  by  f,  the  second  by  |, 
and  the  third  by  ^^,  the  resulting  numbers  will  be  in  arithmetic 
progression.     What  are  the  numbers  ? 

HARMONIC  PROGRESSION 

378.  A  Harmonic  Progression  is  a  series  of  terms  whose 
reciprocals  form  an  arithmetic  progression. 

Thus,  1,  "I",  ^,  I,  ^,  •••  is  a  harmonic  progression,  because  the 
reciprocals  of  the  terms,  1,  3,  5,  7,  9,  •••,  form  an  arithmetic 
progression. 

A  Harmonic  Progression  is  also  called  a  Harmonic  Series. 

3i9.  Any  problem  in  harmonic  progression,  which  is  suscep- 
tible of  solution,  may  be  solved  by  taking  the  reciprocals  of  the 
terms,  and  applying  the  formulae  of  the  arithmetic  progression. 

There  is,  however,  no  general  method  for  finding  the  sum  of 
the  terms  of  a  harmonic  series. 

Ex.  In  the  progression  2,  |,  |,  •••  to  36  terms,  find  the  last 
term. 


PROGRESSIONS  347 

Taking  the  reciprocals  of  the  terms,  we  have  the  arithmetic  progression 

13    5 

2'   2'   2'  ""* 

Here,  a  =  -,  d  =  l,  w  =  36. 

Substituting  in  (I),  §  359,  Z  =  1  +  (36  -  1)  x  1  =  — • 

2 
Then,  —  is  the  last  term  of  the  given  harmonic  series. 

380.  Harmonic  Means. 

We  define  inserting  m  harmonic  means  between  two  numbers,  a 
and  b,  as  finding  a  harmonic  progression  of  m  +  2  terms,  whose 
first  and  last  terms  are  a  and  b. 

Ex.  Insert  5  harmonic  means  between  2  and  —  3. 

We  have  to  insert  5  arithmetic  means  between  -  and 

2  3 

Substituting  a  =  ^,  Z  =  --,  7i  =  7,  in  (I),  §  359, 

-l  =  l  +  6d, --  =  6c?,  or  d  =  -A. 
3     2  6  '  36 

Then  the  arithmetic  series  is    -,   — ,   -,   — , , ,    — -• 

2     36     9     12         18'        36         3 

Therefore,  the  required  harmonic  series  is 

2,   §^,   ^,    12,    -18,    -^,    -3. 
'    13'   2'       '  '         7' 

381.  Let  X  denote  the  harmonic  mean  between  a  and  b. 

Then,  -  is  the  arithmetic  mean  between  -  and  — 
X  a  b 

Then,  by  §  366,  1  =  ^  =  ^,  and  x  =  ^^. 
^  '  X         2  2ab'  a-\-b 

EXERCISE   164 

Find  the  last  terms  of  the  following : 

1.   I,  J/,  ¥?  •••  to  19  terms. 


348 


ALGEBRA 

2. 

-h  -h  -2%. -to  46  terms. 

3. 

-h  -h  -^^  ••'  to  33  terms. 

4. 

IT?  w  TTJ-"to  11  terms. 

5. 

4,  3-8,,  2^  ...to  28  terms. 

6.  Insert  7  harmonic  means  between  —  4  and  i|. 

7.  Insert  8  harmonic  means  between  —  |  and  —  |. 

8.  Insert  6  harmonic  means  between  i  and  —  ^. 


Find  the  harmonic  mean  between : 
9.   land  -|.  .         10.   ^^±^  and  "^ 


d'  +  b' 

11.  Find  the  next  to  the  last  term  of  the  harmonic  progres- 
sion a,  b,  •••to  71  terms. 

12.  If  m  harmonic  means  are  inserted  between  a  and  b,  what 
is  the  third  mean  ? 

13.  The  sixth  term  of  a  harmonic  progression  is  ^,  and  the 
eleventh  term  —  f .     Find  the  fourteenth  term. 

14.  The  geometric  mean  between  two  numbers  is  4,  and  the 
harmonic  mean  ^-.     Find  the  numbers. 

382.  If  any  three  consecutive  terms  of  a  harmonic  series  be 
taken,  the  first  is  to  the  third  as  the  first  minus  the  second  is  to  the 
second  minus  the  third. 

Let  the  terms  be  a,  b,  and  c;  then,  since  -,  -,  and  -  are  in 


arithmetic  progression, 


a    b  c 


1111  b 

=  ---,  or 


c     b      b     a  be  ab 

Multiplying  both  members  by  r^^ — ,  we  have 

b  —  c 

a  _a  —  b 

c      b  —  c 


PROGRESSIONS  349 

383.   Let  A,  G,  and  H  denote  the  arithmetic,  geometric,  and 
harmonic  means,  respectively,  between  a  and  b. 
Then,  by  §§  366,  376,  and  381, 

A  =  ^,  (y  =  V^,  and^=2^^ 


a-\-b 


But,  ^X^  =  a6=(V^)^. 

'  2        a-\-b  ^ 


Whence,         A  x  H=  G\  or  G  =  VA  x  H. 

That  is,  the  geometric  mean  between  two  numbers  is  also  the 
geometric  mean  betiveen  their  arithmetic  and  harmonic  means. 


350  ALGEBRA 


XXVIII.    THE  BINOMIAL  THEOREM 

POSITIVE  INTEGRAL  EXPONENT 

384.  A  Series  is  a  succession  of  terms. 

A  Finite  Series  is  one  having  a  limited  number  of  terms. 
An  Infinite  Series  is  one  having  an  unlimited  number  of  terms. 

385.  In  §§97  and  205  we  gave  rules  for  finding  the  square 
or  cube  of  any  binomial. 

The  Binomial  Theorem  is  a  formula  by  means  of  which  any 
power  of  a  binomial  may  be  expanded  into  a  series. 

386.  Proof  of  the  Binomial  Theorem  for  a  Positive  Integral 
Exponent. 

The  following  are  obtained  by  actual  multiplication : 
(a-\-xy  =  a''^2ax^3^; 

(a  -\- xy  =  a* -\- 4:  a^x  +  6  aV  +  4  cta^  -f  x* ;  etc. 

In  these  results,  we  observe  the  following  laws : 

1.  The  number  of  terms  is  greater  by  1  than  the  exponent 
of  the  binomial. 

2.  The  exponent  of  a  in  the  first  term  is  t^e  same  as  the 
exponent  of  the  binomial,  and  decreases  by  1  in  each  succeed- 
ing term. 

3.  The  exponent  of  x  in  the  second  term  is  1,  and  increases 
by  1  in  each  succeeding  term. 

4.  The  coefficient  of  the  first  term  is  1,  and  the  coefficient  of 
the  second  term  is  the  exponent  of  the  binomial. 

^  5.  If  the  coefficient  of  any  term  be  multiplied  by  the  expo- 
nent of  a  in  that  term,  and  the  result  divided  by  the  exponent 
of  X  in  the  term  increased  by  1,  the  qdotient  will  be  the 
coefficient  of  the  next  following  term. 


THE   BINOMIAL   THEOREM  351 

387.  If  the  laws  of  §  386  be  assumed  to  hold  for  the  expan- 
sion of  (a  +  a?)",  where  n  is  any  positive  integer,  the  exponent 
of  a  in  the  first  term  i^  w,  in  the  second  term  n  —  1,  in  the 
third  term  n  —  2,  in  the  fourth  term  n  —  3,  etc. 

The  exponent  of  x  in  the  second  term  is  1,  in  the  third  term 
2,  in  the  fourth  term  3,  etc. 

The  coefficient  of  the  first  term  is  1 ;  of  the  second  term  n. 

Multiplying  the  coefficient  of  the  second  term,  n,  by  n  —  1, 
the  exponent  of  a  in  that  term,  and  dividing  the  result  by 
the  exponent  of  x  in  the  term  increased  by  1,  or  2,  we  have 

'^y^~   )    as  the  coefficient  of  the  third  term ;  and  so  on. 
1.2 

(A  point  is  often  used  for  the  sign  x  ;  thus,  1  •  2  signifies  1x2.) 
Then,  (a  +  xf  =  a''^  ua^r.^aj  +  '^^^~^]  a"" V 

.         +«(«-l)(r^-2)^.-V+..  (1) 

Multiplying  both  members  of  (1)  by  a-\-x,  we  have 
(a  4-  xy+^ 
=  a-^  +  na^x  +  'l(^^a-y  +  ^(^- l)»-2)^n-v  4. ... 

\  •  Z 

Collecting  the  terms  which  contain  like  powers  of  a  and  a;, 
we  have, 

(a  +  a;)"+^  =  a"+^  +  (n  + 1)  oiTx  +  ["^  ^^  ~  -*-)  +  ^^1  a""  ^ 

ryi(n-l)(n-2)      ^^C^- l)"1^n-2^  .   ... 
[_       1.2.3        ^     1.2    J  ^ 

=  a"+i  +  (ri  4- 1)  a"a;  +  ^i  f^^^  + 1]  «""^^ 

n(n-l)nj^^-1       ,^      ^.,^ 
^     1.2     [    3  J 


352  ALGEBRA 


a^'-'x' 


Then,  (a  +  i»)"+i  =  a«+i  +  (^  + 1)  a"x  +  n  T^^-ti 

,  71  (71  —  1)  Fn  + 1~1   ,,_o  o  , 


=  ay+'  +  (»  +  !)  <t"a;  4-  ("  +  ^)  "•  a- V 

^(„  +  l)n(n-l)^„_y^,.._  (2) 

It  will  be  observed  that  this  result  is  in  accordance  with 
the  laws  of  §  386 ;  which  proves  that,  if  the  laws  hold  for  any 
power  of  ft  +  a;  whose  exponent  is  a  positive  integer,  they  also 
hold  for  a  power  whose  exponent  is  greater  by  1. 

But  the  laws  have  been  shown  to  hold  for  (a  +  xy,  and 
hence  they  also  hold  for  (a  +  xy ;  and  since  they  hold  for 
^     (a  +  xy,  they  also  hold  for  (a  -f-  xy ;  and  so  on. 

Therefore,  the  laws  hold  wfien  the  exponent  is  any  positive 
integer,  and  equation  (1)  is  proved  for  every  positive  integral 
value  of  n. 

Equation  (1)  is  called  the  Binomial  Theorem. 

In  place  of  the  denominators  1-2,  1.2.3,  etc.,  it  is  usual  to  write 
[2,  [3,  etc. 

The  symbol  \n^  read  "factorial-n,"  signifies  the  product  of  the  natural 
numbers  from  1  to  w,  inclusive. 


The  method  of  proof  exemplified  in  §  387  is  known  as  Mathematical 
Induction. 

A  more  complete  form  of  the  proof  of  §  387,  in  which  the  fifth  law  of 
§  386  is  proved  for  any  two  consecutive  terms,  will  be  found  in  §  447. 

388.  Putting  a  =  1  in  equation  (1),  §  387,  we  have 

389.  In  expanding  expressions  by  the  Binomial  Theorem, 
it  is  convenient  to  obtain  the  exponents  and  coefficients  of  the 
terms  by  aid  of  the  laws  of  §  386. 


THE   BINOMIAL   THEOREM  353 

1.  Expand  (a-\-xy. 

The  exponent  of  a  in  the  first  term  is  5,  and  decreases  by  1  in  each 
succeeding  term. 

The  exponent  of  x  in  the  second  term  is  1,  and  increases  by  1  in  each 
succeeding  term. 

The  coefficient  of  the  first  term  is  1  ;  of  the  second,  5. 

Multiplying  5,  the  coefficient  of  the  second  term,  by  4,  the  exponent  of 
a  in  that  term,  and  dividing  the  result  by  the  exponent  of  x  increased  by 
1,  or  2,  we  have  10  as  the  coefficient  of  the  third  term  ;  and  so  on. 

Then,     {a -{■  x)^  =  a^ -\- ^  a%  +  lo  oF'x'^  +  10.  a^x^  +  5  ax*  +  x^. 

It  will  be  observed  that  the  coefficients  of  terms  equally  distant  from 
the  ends  of  the  expansion  are  equal ;  this  law  will  be  proved  in  §  391. 

Thus  the  coefficients  of  the  latter  half  of  an  expansion  may  be  written 
out  from  the  first  half. 

If  the  second  term  of  the  binomial  is  negative,  it  should 
be  enclosed,  negative  sign  and  all,  in  parentheses  before 
applying  the  laws ;  in  reducing,  care  must  be  taken  to  apply 
the  principles  of  §  96. 

2.  Expand  (1  -  x)\ 

(l-a:)6  =  [H-(-x)]6 

=  16^6.15.  (_a;)  +  ,1^.14.  (^a:)2  +  20.l3.  {-xY 

+  15.12.  (-x)*  + CI.  (-x)5  +  (-x)6 
=  1  -  6  X  +  16  x2  -  20  x3  +  15  X*  -  6  x5  +  a;6. 

If  the  first  term  of  the  binomial  is  an  arithmetical  number,  it  is  con- 
venient to  write  the  exponents  at  first  without  reduction ;  the  result 
should  afterwards  be  reduced  to  its  simplest  form. 

If  either  term  of  the  binomial  has  a  coefficient  or  exponent 
other  than  unity,  it  should  be  enclosed  in  parentheses  before 
applying  the  laws. 

3.  Expand  (3  m2--v^^y. 

(3  w2  -  ^ny  =  [ (8  7^2)  +  (  _  n^)  ]4 

=  (3  m'^y  +  4(3  m2)3( _  n^)  +  6(3  m2)2(_  n^y 

+  4(3  m2)  (  -  n^y  +  ( -  n^y 
=  81  »n8  -  108  mHi^  +  54  m*>i*  -  12  m^n+  n^. 


354  ALGEBRA 

f!' 

A  trinomial  may  be  raised  to  any  power  by  the  Binomial 
Theorem,  if  two  of  its  terms  be  enclosed  in  parentheses,  and 
regarded  as  a  single  term ;  but  for  second  powers,  the  method 
of  §  204  is  shorter. 

4.   Expand  (x--2x-2y. 
(a;2  _  2  X  -  2)*  =  [(x2  -  2  x)  +  (- 2)]* 

=  (a:2  -  2  x)*  +  4(a;2  -  2  a;)3(  -  2)  +  6(x2  -  2  a;)2(  -  2)2 

+  4(^2  _  2  x) (-  2)3  +  (-  2)* 
=  x^  -  S x^ -\-2i x^  -  S2x^ -h  16 x^ 

-8(x6-6x6  +  12x*-8a;3) 
+  24(a;4  -  4  x^  +  4  a;2)  -  32(x2  -2x)+16 
=  x8-8  xH  16  a;6  +  16x5-56  ic*-32  ic3+64  x2  +  64  x+ie. 


EXERCISE  165 

Expand  the  following : 

1.  (a-\-xy.  13^  (x^^sy. 


21. 


2.    (n-hiy 


14.  (i-x^y. 


(f-'-J- 


3.  a-yy.  '  _  '     ^22.  (sx^ — ^y 

4.   (a -a;)'.  15.    (Va ^Y.  ^  2  kV 


5.  (:^y  +  ^y.  ^  ^'^'^         23.   (Zai  +  Vaf. 

6.  (x  +  2  2,)'.  16.   (mS  +  4«-})^        24.  fsVcP-^Y 

7.  (2-ay.     17.  r.*,-i+iY.   .^  X,-  :S. 


8.  (3a^-6^)^  ^  ^^  25.    ^^Va;-^J. 

9.  (a-4-a^'^)^  18.  M  +  ^fY.                .2m^^  .M^ 

10.  {2x^^fy.  ^  ^^  ^^-   ^^  +  3^J* 

11.  («  +  .)«  19.  (5.--^ny.  ^^     /^-__1_V 

12.  (a-2H-</^)6.  20.  (2a:^  +  2/"^)'.              "    V           Zi/hV 

28.    (a-6)^  29.  {a  +  iy\ 

30.  (iB2^a,^iy^  32_  (a^_3a._i)4^  34.   (i-^x^o?)'. 

31.  (2-a;  +  a^)*.  33.  {^x'  +  x-^y.  35.    (a;-'  +  .T-3)^ 


THE   BINOMIAL   THEOREM  355 

390.    To  find  the  rtli  or  general  term  in  the  expansion  of(a-\-xY. 

The  following  laws  hold  for  any  term  in  the  expansion  of 
(a-\-xY,  in  equation  (1),  §  387 : 

1.  The  exponent  of  x  is  less  by  1  than  the  number  of  the 
term. 

2.  The  exponent  of  a  is  n  minus  the  exponent  of  x. 

3.  The  last  factor  of  the  numerator  is  greater  by  1  than  the 
exponent  of  a. 

4.  The  last  factor  of  the  denominator  is  the  same  as  the 
exponent  of  x. 

Therefore  in  the  rth  term,  the  exponent  of  x  will  be  r  —  l. 
The  exponent  of  a  will  be  n  —  (?-  —  1),  or  n  —  r  -\-l. 
The  last  factor  of  the  numerator  will  be  n  —  r  -\-  2. 
The  last  factor  of  the  denominator  will  be  r  —  1. 
Hence,  the  rth  term 

^  n(n  -  1)  (n  -  2)  . . .  (n  -  r  +  2)     n-r+ W-l  QN 

1.2.3...  (r-1)  *  ^^ 

In  finding  any  term  of  an  expansion,  it  is  convenient  to  obtain 
the  coefficient  and  exponents  of  the  terms  by  the  above  laws. 

Ex.   Find  the  8th  term  of  (3  a^  -  h-y\ 

We  have,  (3  a^  -  h-^y^  =  [(3  a^)  +  (-  6-i)]ii. 

In  this  case,  n  =  11,  r  =  8. 

The  exponent  of  (—  &~i)  is  8  —  1,  or  7. 

The  exponent  of  (3  a^)  is  11  —  7,  or  4. 

The  first  factor  of  the  numerator  is  11,  and  the  last  factor  4  +  1,  or  5. 

The  last  factor  of  the  denominator  is  7. 

Then,  the  8th  term  =  12  •  10  •  9  •  8  -  7  •  6  .  5^3  ^^y^^-  b-^y 
1.2.3.4.5.6.7^  ^  ^ 

=  330(81  a2)  (  _  5-7)  =  _  26730  a^ft-T. 

If  the  second  term  of  the  binomial  is  negative,  it  should  be  enclosed, 
sign  and  all,  in  parentheses  before  applying  the  laws. 

If  either  term  of  the  binomial  lias  a  coefficient  or  exponent  other  than 
unity,  it  should  be  enclosed  in  parentheses  before  applying  the  laws. 


356 


ALGEBRA 


Find  the : 

1.  4th  term  of  (a  -f-  xj. 

2.  6th  term  of  {n  +  iy\ 

3.  5th  term  of  (a  -  hf. 

4.  7th  term  of  (1  -  a^. 

5.  8th  term  of  (aj*  +  fY^- 

6.  5th  term  of  (a^  +  2  x~^y\ 

7.  9th  term  of  {x^  -  x'^y^ 

8.  10th  term  of  (^^'  +  -Y'. 

\b       aj 

14.  8th  term  of 

15.  Middle  term 


EXERCISE  166 

9.   10th  term  of /"-v^m^-^y'. 
10.   6th  term  of  (x'  -  4  yfy'. 


11.    Tthtermof  (^m~*  +  ^^^^ 


12. 

4th  term 

of  (m-* 

—  5  mny^. 

13. 

9th  term  of  fa^+ 

1  Y^ 

4^ 

(^■-W 

of  (^3  a' 


+ 


6*Y' 
2;  * 


391.  Multiplying  both  terms  of  the  coefficient,  in  (1),  §  390, 
by  the  product  of  the  natural  numbers  from  1  to  n  —  r  +  1, 
inclusive,  the  coefficient  of  the  rth  term  becomes 


n(n-l)  ...{n-r-\-2)'(n-r-hl)  '"2  '1  _ 
I  r-1  xl  -2...  (n-r  +  1) 


\n 


r+1 


Since  the  number  of  terms  in  the  expansion  is  n  + 1,  the  rth 
term  from  the  end  is  the  (n  —  r-\-  2)th  from  the  beginning. 

Then,  to  find  the  coefficient  of  the  ?*th  term  from  the  end,  we 
put  in  the  above  formula  n  —  r  +  2  for  r. 

Then,  the  coefficient  of  the  rth  term  from  the  end  is 

I  71  \  n 

' Qp      ' . 

I  yt  —  r  +  2  —  1  I  71  — (yi  — r  +  2)4-l'        |  n  —  r  + 1  |  r  — 1 

Hence,  in  the  expansion  of  (a  +  a?)",  the  coefficients  of  terms 
equidistant  from  the  ends  of  the  expansion  are  equal. 


UNDETERMINED  COEFFICIENTS 


367 


XXIX.    UNDETERMINED  COEFFICIENTS 

392.  Infinite  Series  (§  384)  may  be  developed  by  Division, 
or  by  Evolution. 

Let  it  be  required,  for  example,  to  divide  1  by  1  —  a;. 

l-a;)l(l  +  x  +  a^  +  - 
1-a; 

X 


Then, 


1-x 


l  +  a;  +  ar^  +  a^  + 


(1) 


Again,  let  it  be  required  to  find  the  square  root  of  1  +  a;, 
l  +  a; 


^2      8^ 


2  + 


X  + 


2-{-x 


x" 


X     y? 


Then,  V^^:^  =  l  +  |-|  + 


(2) 


It  should  be  observed  that  the  series,  in  (1)  and  (2),  do  not  give  the 
values  of  the  first  members  for  every  value  of  x ;  thus,  if  x  is  a  very 
large  number,  they  evidently  do  not  do  so. 


EXERCISE  167 
Expand  each  of  the  following  to  four  terms : 


1. 


2. 


3  +  4a; 
l  +  2a;* 


3. 


4a; 


2  +  6a;-a^ 


5.    VTT6^. 


4     2  +  4a;-5a^,  g     Vr32^. 

3_6a^  +  7a^ 


358  ALGEBRA 


7.  VT+a.  9.    Va^  +  a!2/  +  2/'-         H-    va^^  +  l. 

8.  VI -5a.  10.    V9a2+6.  12.    A/a«^3  63. 

CONVERGENCY  AND  DIVERGENCY  OF  SERIES 

393.  An  intinite  series  is  said  to  be  Convergent  when  the  sum 
of  the  first  ?i  terms  approaches  a  fixed  finite  number  as  a  limit 
(§  318),  when  n  is  indefinitely  increased. 

An  infinite  series  is  said  to  be  Divergent  when  the  sum  of  the 
first  n  terms  can  be  made  numerically  greater  than  any  assigned 
number,  however  great,  by  taking  n  sufticiently  great. 

394.  Consider,  for  example,  the  infinite  series 

1 -\- X -\- oir^ -{- a^ -] . 

I.  .Suppose  X  =  Xi,  where  osi  is  numerically  <  1. 
The  sum  of  the  first  n  terms  is  now 

1  +  0^1  +  o^i^  +  -  +  xr'  =  \=^^  (§  103). 

1  — iCj 

If  n  be  indefinitely  increased,  x^"  decreases  indefinitely  in 
absolute  value,  and  approaches  the  limit  0. 

Then  the  fraction approaches  the  limit 


1  —  i»l  1  —  £Cl 

That  is,  the  sum  of  the  first  n  terms  approaches  a  fixed  finite 
number  as  a  limit,  when  n  is  indefinitely  increased. 

Hence,  the  series  is  convergent  when  x  is  numerically  <  1. 

II.  Suppose  x  =  l. 

In  this  case,  each  term  of  the  series  is  equal  to  1,  and  the 
sum  of  the  first  n  terms  is  equal  to  n;  and  this  sum  can  be 
made  to  exceed  any  assigned  number,  however  great,  by  taking 
n  sufficiently  great. 

Hence,  the  series  is  divergent  when  x  =  l. 

III.  Suppose  £c  =  —  1. 

In  this  case,  the  series  takes  the  form  1  — 1  +  1  —  l4-.'-,  and 
the  sum  of  the  first  n  terms  is  either  1  or  0  according  as  n  is 
odd  or  even. 


UNDETERMINED  COEFFICIENTS  359 

Hence,  the  series  is  neither  convergent  nor  divergent  when 

An  infinite  series  which  is  neither  convergent  nor  divergent 
is  called  an  Oscillating  Series. 

IV.   Suppose  X  =  Xi,  where  x^  is  numerically  >  1. 
The  sum  of  the  first  n  terms  is  now 

1  +  0^1  +  a^i^  +  -  +  X,--'  =  ^^  (§  103). 

By  taking  n  sufficiently  great,  — —  can  be  made  to  numeri- 

cally  exceed  any  assigned  number,  however  great. 

Hence,  the  series  is  divergent  when  x  is  numerically  >  1. 

395.   Consider  the  infinite  series 

l  +  x-{-x^-\-x^-\ , 

developed  by  the  fraction (§  392). 

1  —  X 

Let  x=  .1,  in  which  case  the  series  is  convergent  (§  394). 

The  series  now  takes  the  form  1  +  .1  +  .01  +  .001  H ,  while 

the  value  of  the  fraction  is  — ,  or  —  • 

In  this  case,  however  great  the  number  of  terms  taken,  their 
sum  will  never  exactly  equal  Y- 

But  the  sum  approaches  this  value  as  a  limit ;  for  the  series 
is  a  decreasing  geometric  progression,  whose  first  term  is  1,  and 

ratio  .1 ;  and,  by  §  373,  its  sum  to  infinity  is -,  or  —  • 

JL  —  .J.  «7 

Thus,  if  an  infinite  series  is  convergent,  the  greater  the  num- 
ber of  terms  taken,  the  more  nearly  does  their  sum  approach 
to  the  value  of  the  expression  from  which  the  series  was 
developed. 

Again,  let  x  =  10,  in  which  case  the  series  is  divergent. 

The  series  now  takes  the  form  1 -f  10 +  100  +  1000  + •••, 

while  the  value  of  the  fraction  is -— ,  or  — -• 

1-10  9 


360  ALGEBRA 

In  this  case  the  greater  the  number  of  terms  taken,  the 
more  does  their  sum  diverge  from  the  value  —  ^. 

Thus,  if  an  infinite  series  is  divergent,  the  greater  the  number 
of  terms  taken,  the  more  does  their  sum  diverge  from  the  value 
of  the  expression  from  which  the  series  was  developed. 

It  follows  from  the  above  that  an  infinite  series  cannot  be 
used  for  the  purposes  of  demonstration,  if  it  is  divergent. 

THE  THEOREM  OF  UNDETERMINED  COEFFICIENTS 

396.  An  important  method  for  expanding  expressions  into 
series  is  based  on  the  following  theorem : 

If  the  series  A  -f-  Bx  -\-  Cx^  +  Da?  -\ is  always  equal  to  the 

series  A'  +  B'x  +  C'x^  +  D'x^  +  •  •  •,  when  x  has  any  value  which 
makes  both  series  convergent,  the  coefficients  of  like  powers  of  x  in 
the  series  will  be  equal;  that  is,  A=A',  B  =  B',  C=  C,  etc. 

For  the  equation 

A-i-Bx+Cx'-{-Da?  +  '"=A'  +  B'x+C'a?-\-D'a?+'"     (1) 

is  satisfied  when  x  has  any  value  which  makes  both  members 
convergent. 

But  both  members  are  convergent  when  x  =  0]  for  the  sum 

of  all  the  terms  of  the  infinite  series  a -^  bx -{- ca? -^  dx^ -i 

is  equal  to  a  when  x  =  0. 

Then,  the  equation  (1)  is  satisfied  when  x  =  0. 

Putting  a;  =  0,  we  have  A  =  A'. 

Subtracting  A  from  the  first  member  of  the  equation,  and  its 
equal  A'  from  the  second  member,  we  obtain 

Bx  +  Cx^-^Da?-^'"=B'x-\-C'x'-\-D'a?+"'.  (2) 

Dividing  each  term  by  x, 

B-^-Cx  +  Dx^-h''.  =B'  -h  C'x-j-D'x'+'".  (3) 

This  equation  also  is  satisfied  when  x  has  any  value  which 
makes  both  members  convergent ;  and  putting  a;  =  0,  we  have 

B  =  B'. 


UNDETERMINED  COEFFICIENTS  361 

In  like  manner,  we  may  prove  (7=  C",  D  =  D',  etc. 

The  proof  of  §  396  is  open  to  objection  iu  one  respect. 

We  know  that  (2)  has  the  same  roots  as  (1),  including  the  root  0;  but 
when  we  divide  by  z,  all  that  we  know  about  the  resulting  equation  is 
that  it  has  the  same  roots  as  (2) ,  except  the  root  0. 

Thus,  we  do  not  know  that  0  is  a  root  of  (3) ,  though  we  assume  it  in 
proving  that  B  =  B'. 

A  more  rigorous  proof  of  the  Theorem  of  Undetermined  Coefficients 
will  be  found  in  §  450. 

397.  The  theorem  of  §  396  holds  when  either  or  both  of  the 
given  series  are  finite. 

EXPANSION  OF  FRACTIONS 

2  _  3aj2  _  gj3 

398.  1.   Expand — —  in  ascending  powers  of  x. 

1  — 2a;  +  3ar 

Assume      ^  ~^^^  ~^\=  A-]-  Bx  +  Cx'^  +  Dx^  +  Ex^  +  ■«.,  (1) 

where  A^  J5,  O,  Z>,  E^  •••,  are  numbers  independent  of  x. 

Clearing  of  fractions,  and  collecting  the  terms  in  the  second  member 
involving  like  powers  of  x,  we  have 


2_3x2-a;3  =  ^+    B\x+     C 

-2^1     -2B 

+  3^ 


-20 
+  35 


x^-^    E 
-2D 

+  30 


a^  +  ....       (2) 


A  vertical  line,  called  a  6ar,  is  often  used  in  place  of  parentheses. 

Thus,  +    B\x  \&  equivalent  to  (J5  —  2  A)x. 

-2a\ 

The  second  member  of  (1)  must  express  the  value  of  the  fraction  for 
every  value  of  x  which  makes  the  series  convergent  (§395);  and  there- 
fore equation  (2)  is  satisfied  when  x  has  any  value  which  makes  the 
second  member  convergent. 

Then,  by  §  397,  the  coefficients  of  like  powers  of  x  in  (2)  must  be 
equal ;  that  is, 

A=     2. 

B-2A=     0;  or,  5  =  2^  =4. 

(7-25  +  3.A  =  -3;  or,  O  =  2  J5  -  3^  -  3  =- 1. 

I>-20+35=-l;  or,  i>  =  20-35-l=-15. 

^-21>  +  30=      0;or,  J?  =  2D-30         =-27;  etc. 


assume 


362  ALGEBRA 

Substituting  these  values  in  (1),  we  have 

2-3a;2-x8  ^  2  +  4  x  -  a:2  -  15x8  -  27a^ . 

l-2x  +  3x2 

The  result  may  be  verified  by  division. 

The  series  expresses  the  value  of  the  fraction  only  for  such  values  of  x 
as  make  it  convergent  (§  395) . 

If  the  numerator  and  denominator  contain  only  even  powers 
of  X,  the  operation  may  be  abridged  by  assuming  a  series  con- 
taining only  the  even  powers  of  a;. 

Thus,  if  the  fraction  were  — — — - — — — ,  we  should 

1  —  3  or -\- 5  x'^ 

it  equ al  to  ^  +  5a^  +  Cx'  -\-Dx^  +  Ea^+"'. 

In  like  manner,  if  the  numerator  contains  only  odd  powers 
of  X,  and  the  denominator  only  even  powers,  we  should  assutne 
a  series  containing  only  the  odd  powers  of  x. 

If  every  term  of  the  numerator  contains  x,  we  may  assume  a 
series  commencing  with  the  lowest  power  of  x  in  the  numerator. 

If  every  term  of  the  denominator  contains  x,  we  determine 
by  actual  division  what  power  of  x  will  occur  in  the  first  term 
of  the  expansion,  and  then  assume  the  fraction  equal  to  a  series 
commencing  with  this  power  of  x,  the ,  exponents  of  x  in  the 
succeeding  terms  increasing  by  unity  as  before. 

2.    Expand  — — — —  in  ascending  powers  of  x. 

O  Xi    —  Xj 

Dividing  1  by  3  x^,  the  quotient  is  — - ;  we  then  assume, 

o 

^        -.4x-2  +  5x-i  +  0+i)x  +  ^x2+-..  (3) 


3  X2  -  X8 

Clearing  of  fractions, 

l  =  3^  +  35|x  +  3C|x2  +  3i)|x3 +  3^1x4+  -. 

-  a\    -  b\    -    c\    -  b\ 

Equating  coefficients  of  like  powers  of  x, 

3^=1,  35-^  =  0,  30-5  =  0,  32)- e  =  0,  3^-D  =  0;  etc. 

Whence,       A=\   B=.\    C  =  — ,   D  =  — ,    iE'  =  — ,   etc. 
3  9  27  81'  243' 


UNDETERMINED  COEFFICIENTS 


363 


Substituting  in  (3) , 


3x2 


T      "9"     27      81      243 


In  Ex.  1,  ^=22)  —  3C;  that  is,  the  coeflBcient  of  x*  equals  twice  the 
coeificient  of  the  preceding  term,  minus  three  times  the  coefficient  of  the 
next  but  one  preceding. 

It  is  evident  that  this  law  holds  for  the  succeeding  terms ;  thus,  the 
coefiBcient  of  x^  is  2  x  (-  27)-  3  x  (-  15),  or  -  9. 

After  the  law  of  coefficients  has  been  found  in  any  expansion,  the  terms 
may  be  found  more  easily  than  by  long  division  ;  and  for  this  reason  the 
method  of  §  398  is  to  be  preferred  when  a  large  number  of  terms  is 
required. 

The  law  for  Ex.  2  is  that  each  coefficient  is  one-third  the  preceding. 


EXERCISE  168 


Expand  each  of  the  following  to  five  terms  in  ascending 
powers  of  x\ 


1. 


2. 


3. 


5. 


3  +  2a; 
\-x  ' 

l-6x 

4  +  a;^ 
l-Sx"' 

2x 

'S-Bx^' 

l-\-4:X  —  xr 
l_a;-f  3a^* 


g     2-x-{-3x^  jj         3  +  4a^ 


9. 


10. 


1+20^2 

l-4a.-2 

l^4.x-x' 

X  — 5x^  —  7  x* 

\-2x-4.x' 

x^-3^ 

^^x-2^ 

4  +  a;-5a^ 

12. 


13. 


2-aj  +  3a^ 


15. 


2-f3a^-a^ 

3 
2a^  +  a;^* 
2Jf.^x-Sx\ 
a^  —  5  ic^  +  ic^ 


^^     l-6a^  + 4a^ 
X'\-x^-2^  ' 


l-2a;  +  3a^ 

2a^  +  4:X^-\-a^' 


EXPANSION  OF   SURDS 


399.  Ex.    Expand  v  1  —  a;  in  ascending  powers  of  x. 

Assume       y/l  -  x  =  A  +  Bx  -\-  Cx^  +  Dx^  +  Ex^  +  .... 


(1) 


Squaring  both  members,  we  have  by  §  204, 


1  -  X  =  ^2 


-\-2AB 


x+      B^ 

+  2AG 


+  2  AD 
+  2BC 


X3+  (72 

+  2AE 
+  2BD 


x*  + 


364  ALGEBRA 

Equating  coefficients  of  like  powers  of  cc, 

A^=      1  ;   or,  ^  =  1. 

2^5  =  - 1;   or,  ^=_J_=_1. 
2A         2 

B^-\-2AC=     0;   or,   (7  =  -  — =  -1. 

2  A  o 

2AD  +  2BC=     0;  or,  D=-^=-l. 

A  16 

C^  +  2AE  +  2BD=     0:   or,  ^  =  - .^1+^^^  =  _  A-  etc. 

2  A  128' 

Substituting  these  values  in  (1),  we  have 

Vnr^^i_?_^_^_6^ 

2      8      16      128 

The  result  may  be  verified  by  Evolution. 

The  series  expresses  the  value  of  Vl  —  a;  only  for  such  values  of  x  as 
make  it  convergent. 

EXERCISE  169 

Expand  each  of  the  following  to  five  terms  in  ascending 
powers  of  x: 

1.  Vl  +  2a;.  3.    Vl-4iB-har^.  5.    -s/l  +  dx. 

2.  VI -3a;.  4.    -Vl-^x-x",  6.    </l-x-2x'. 


PARTIAL  FRACTIONS 

400.  If  the  denominator  of  a  fraction  can  be  resolved  into 
factors,  each  of  the  first  degree  in  x,  and  the  numerator  is  of  a 
lower  degree  than  the  denominator,  the  Theorem  of  Undeter- 
mined Coefficients  enables  us  to  express  the  given  fraction  as 
the  sum  of  two  or  more  partial  fractions^  whose  denominators 
are  factors  of  the  given  denominator,  and  whose  numerators 
are  independent  of  x. 

401.  Case  I.     No  factors  of  the  denominator  equal. 

1.   Separate  — -^-J^t into  partial  fractions. 

^  (3a;-l)(5a;  +  2)  ^ 


UNDETERMINED  COEFFICIENTS  365 

^^"°"'  (3x-l)(6»  +  2)=3^3T  +  5^Ti'  ^^^ 

where  ^  and  £  are  numbers  independent  of  x. 

Clearing  of  fractions,      19  x  +  1  =  ^(5  a;  +  2)  +  .5(3  x-V). 

Or,  19  X  +  1  =  (5  ^  +  3  LB)a;  +  2  ^  -  ^.  (2) 

The  second  member  of  (1)  must  express  the  value  of  the  given  fraction 
for  every  value  of  x. 

Hence,  equation  (2)  is  satisfied  by  every  value  of  x  ;  and  by  §  397,  the 
coefBcients  of  like  powers  of  x  in  the  two  members  are  equal. 

That  is,  5  ^  +  3  5  =  19, 

and  2A-    B  =  \. 

Solving  these  equations,  we  obtain  J.  =  2  and  ^  =  3. 

Substituting  in  (1),  ,„     -^^^^.t  ^     ^^  =  ^-^  +  r^-Z' 

^  ^    (3x-l)(5x  +  2)      3x-l      5x  +  2 

The  result  may  be  verified  by  finding  the  sum  of  the  partial 
fractions. 

2.    Separate  — ^-i- into  partial  fractions. 

The  factors  of  2  x  -  x^  -  x^  are  x,  1  -  x,  and  2  +  x  (§  116). 

Assume  then        ^     "^ +/       =^+_A-  +  ^-. 
2x-x2-x8      X      1-x     2  +  x 

Clearing  of  fractions,  we  have 

X  +  4  =  ^(1  -  X)  (2  +  X)  +  5x(2  +  x)  +  Ox  (1  -  X). 

This  equation,  being  satisfied  by  every  value  of  x,  is  satisfied  when  x  =  0. 
Putting  X  =  0,  we  have  4  =  2^,  or  Jl  =  2. 
Again,  the  equation  is  satisfied  when  x  =  1. 

Putting  X  =  1,  we  have  5  =  3  5,  or  J5  =  — 

o 

The  equation  is  also  satisfied  when  x  =  —  2. 

Putting  X  =  -  2,  we  have  2  =  -  6  C,  or  O  =  -  -• 

3 

5  _1 

Then,    _^+^  =  ?  + J_  +  ^^2^        ' 


2x-x2-x8     x"^l-x"^2  +  x     x3(l-x)      3  (2+x) 


366  ALGEBRA 

To  find  the  value  of  A,  in  Ex.  2,  we  give  to  x  such  a  value  as  will  make 
the  coefficients  of  B  and  C  equal  to  zero  ;  and  we  proceed  in  a  similar 
manner  to  find  the  values  of  B  and  C. 

This  method  of  finding  J.,  B,  and  G  is  usually  shorter  than  that  used 
in  Ex.  1. 

EXERCISE  170 
Separate  the  following  into  partial  fractions : 

1.     - — :; r*  «*•     -^ t^ — *  0- 


9iB2_4  x3-16a;  ^  ^- 3  ax"  -  4.  o?x 

23a:4-25  ^  6a;-ll  ^  10-9a; 


6a;2  +  5a;  6a^4-13aj  +  6  5a^-14a;  +  8 

12  +  17(c-2a;2  g      4  +  14a;-2a;2 


(1  +  3  a;).(9  -h  6  a;  -  8  a:2>^  (a^-5  a;)(a^-4) 

402.   Case  II.     All  the  factors  of  the  denominator  equal. 

Let  it  be  required  to  separate   — -p^ —   into  partial 

fractions.  \^~   ) 

Substituting  2/  +  3  for  «,  the  fraction  becomes 

(y  +  3y-ll(y  +  3)  +  26^y^-5y  +  2^1      5  ^  2 

f  f  y    y^    f' 

Keplacing  2/  by  a;  —  3,  the  result  takes  the  form 

_J____5__     _2_ 
x-3     (x-Zy^  {x-^Y 

This  shows  that  the  given  fraction  can  be  expressed  as  the 
sum  of  three  partial  fractions,  whose  numerators  are  indepen- 
dent of  X,  and  whose  denominators  are  the  powers  of  ic  — 3 
beginning  with  the  first  and  ending  with  the  third. 

Similar  considerations  hold  with  respect  to  any  example 
under  Case  II;  the  number  of  partial  fractions  in  any  case 
being  the  same  as  the  number  of  equal  factors  in  the  denomi- 
nator of  the  given  fraction. 

Ex.     Separate  — — ^^— -  into  partial  fractions. 
^  (3a;  +  5)2  ^ 


UNDETERMINED   COEFFICIENTS  367 

In  accordance  with  the  above  principle,  we  assume  the  given  fraction 
equal  to  the  sum  of  two  partial  fractions,  whose  denominators  are  the 
powers  of  3  x  +  5  beginning  with  the  first  and  ending  with  the  second. 

Thatis,  6^  +  5    ^_A_+         B 


(3  X  +  5)'-^     3  a;  +  5      (3  x  +  5)2 
Clearing  of  fractions,  6x  +  5  =  J.(3a:4-  5) +  5. 

=  ^Ax  +  bA-^  B. 
Equating  coefficients  of  like  powers  of  x, 
3^  =  6, 
and  bA-\-B  =  b. 

Solving  these  equations,  A  =  2  and  5  =  —  5. 
Whence,  6x  +  5    _      2  5 


(3x+5)2     3x  +  5      (3x  +  5)2 

EXERCISE  171 
Separate  the  following  into  partial  fractions : 

J  24  a; +  2  3     12^-f  7a;-l  .     16a^-19 

^x'  +  TZx^^       '        (l  +  3xf     '  '    {4:x-3y' 

2     2fl^-lla;  +  3      ^     6x'  +  12x-10        g     a^-2af-7x 

(x-4:y    '     '      (s-j-2xy    *      *      (x-{-iy 

y     2a^-18x'-{-24:X-15  «     18  a;  +  54  cc^  +  27  af 

(a;-2)^  '  '  (2  +  3  «)* 

403.   Case  III.    Some  of  the  factors  of  the  denominator  equal. 

Ex.    Separate  — ^-^  into  partial  fractions. 

X  \X  -j—  J. ) 

The  method  in  Case  III  is  a  combination  of  the  methods  of  Cases  I  and 

II ;  we  assume, 

a;2  _  4  a;  +  3  ^  ^        B  G 

X{x+iy  X       X+1        (X+l)2' 

Clearing  of  fractions, 

x2  _  4  X  +  3  =  ^(x  4-  1)2  -f  Bx{x  +  1)  +  Cx 

=  (^  +  S)x2  +{2A  +  B-\-Cyc-\-A. 


368  ALGEBRA 

Equating  coefficients  of  like  powers  of  x, 
A-\-B  =  l, 
2A  +  B-\-C  =  -4, 
and  ^  =  3. 

Solving  these  equations,  A  =  S,  B  =  —  2^  and  (7  =  —  8. 
Whence,  ^^-4x  +  3_3        2  8 


x{x  +1)2       X     a:  +  1      (x  +  I)* 

The  following  general  rule  for  Case  III  will  be  found  convenient : 

A  fraction  of  the  form should  be  assumed 

equal  to  (x  + a)(x  + 6)  ...  (x  +  m)'•... 


x  +  a     x-\-h  x  +  m     {x  +  my  (x  +  my 

Single  factors  like  x  -\-  a  and  x  +  &  having  single  partial  fractions  cor- 
responding, arranged  as  in  Case  I ;  and  repeated  factors  like  (x  +  my 
having  r  partial  fractions  corresponding,  arranged  as  in  Case  II. 


EXERCISE    172 
Separate  the  following  into  partial  fractions  : 

x(x+3f    '  '  a?(x-\-lf 

o     3a^  +  7a^  +  24aj-16  5     -4a;3^  29  a^-36  a;-9 


a?(x-^)  x{x-l){x-^) 

3       14aj2-53a;-4  ^  l-lSx-4.x' 


(3  a;  +  2)  (2  a;  -  3)2  (8  aj2  -  2  a;  -  3)  (2  x  + 1) 

404.  If  the  degree  of  the  numerator  is  equal  to,  or  greater 
than,  that  of  the  denominator,  the  preceding  methods  are 
inapplicable. 

In  such  a  case,  we  divide  the  numerator  by  the  denominator 
until  a  remainder  is  obtained  which  is  of  a  lower  degree  than 
the  denominator. 

Ex.    Separate  — — into  an  integral  expression  and 

partial  fractions.  ~  ^ 


UNDETERMINED   COEFFICIENTS  369 

Dividing  a^  —  3  x^  —  1  by  x^  —  x,   the  quotient  is  x  —  2,  and  the  re- 
mainder   —  2  X  —  1 ;  we  then  have 

^       »''-3^'-'=»-2  +  -^^-^-  (1) 

X2  —  X  X'-^  -  X 

We  can  now  separate  ~     ^  ~      into  partial  fractions  by  the  method 

X''^  —  X 

1         3 
of  Case  I  :  the  result  is 


Substituting  in  (1),  — — '— ^  =  x  -  2  + — 

X'^  -  X  X       X  —  1 

Another  way  to  solve  the  above  example  is  to  combine  the  methods  of 
398  and  401,  and  assume  the  given  fraction  equal  to 

Ax  +  B  +  ^+    ^ 


X        X  —  1 


EXERCISE  173 


Separate  each  of  the  following  into  an  integral  expression 
and  two  or  more  partial  fractions  : 


1. 


(a;-2)(3a;  +  l)  *  :>^(x-l) 


(a;  4-3)^  '  *  x'ix  +  lf 

5     2a^-8a;^  +  2a;^-5a^-fl2a;^-fl;  +  4    ' 

iC^(£C— 4) 

405.  If  the  denominator  of  a  fraction  can  be  resolved  into 
factors  partly  of  the  first  and  partly  of  the  second,  or  all  of  the 
second  degree,  in  x,  and  the  nurnerator  is  of  a  lower  degree 
than  the  denominator,  the  Theorem  of  Undetermined  Coeffi- 
cients enables  us  to  express  the  given  fraction  as  the  sum  of 
two  or  more  partial  fractions,  whose  denominators  are  factors 
of  the  given  denominator,  and  whose  numerators  are  inde- 
pendent of  X  in  the  case  of  fractions  corresponding  to  factors 
of  the  first  degree,  and  of  the  form  Ax  +  B  in  the  case  of 
fractions  corresponding  to  factors  of  the  second  degree. 


370  ALGEBRA 

The  only  exceptions  occur  when  the  factors  of  the  denominator  are  of 
the  second  degree  and  all  equal. 

Ex.    Separate  — — -  into  partial  fractions. 

3/   -p  JL 
The  factors  of  the  denominator  are  cc  +  1  and  x"^  —  x  +  1. 

Assume  then  -Ji—  =  -A_  +    Bx+C  ,j. 

x^-\-lx  +  lx^-x-\-l  ^  ^ 

Clearing  of  fractions,     1  =  A(x'^  -  x  4-  1)  +  {Bx  +  C)  (a;  +  1). 

Or,  lz={A  +  B)x'^  +  {-A  +  B  +  C)x-\-A-\-C. 

Equating  coefficients  of  like  powers  of  cc, 

A-\-  B  =  0, 

-J.  +  5  +  C  =  0, 

and  A-\-C=l. 

1  1  2 

Solving  these  equations,  ^  =  - ,   B  =  — ,   and  G  =  — 

3  3  3 

Substituting  in  (1),  —1-  = ^^^ 

EXERCISE    174 

Separate  the  following  into  partial  fractions : 

2705^  +  8 

x'-l 
6    3-8a;-4a^ 

REVERSION  OF  SERIES 

406.   To  revert  a  given  series  y  =  a-\- bx"^  +  ca;"  +  '•  •  •  is  to  ex- 
press ic  as  a  series  proceeding  in  ascending  powers  of  y. 

Ex.     Eevert  the  series  y  =  2x~3x^-{-4:X^  —  5x^-{-  •••. 

Assume  x  =  Ay  +  By^ -^Ci/ +  Dy^  + -.  (1) 


3a^_4a;-h4 

a^-1 

104-3aj- 

-llaj2 

{Sx-2)(a^- 

-x-i-2) 

x'-hSx- 

5 

UNDETERMINED  COEFFICIENTS 


371 


Substituting  in  this  the  given  value  of  y, 

x  =  A(2x-Sx'^-\-ix^-bx^+  •••) 

+  ^(4  a;2  4- 9  a;4  -  12  x3  +  16  a^  +  ...) 

4- 0(8x3 -36aj4+...)  +  l>(16x* +...)+ — 


That  is, 


2AX-SA 
4-45 


x2+   4A 

x^-   6  A 

-12J5 

+  25  5 

+   SC 

-36(7 

+  16i> 

X4  + 


Equating  coeflBcients  of  like  powers  of 

X, 

2A  = 

=  1; 

-3^+45= 

:0: 

4A 

-125  +  80  = 

:0; 

-5A+2bB- 

-360+16D  = 

:0; 

etc. 

Solving,                    A  = 

--,  5--,  0- 

.  5 
'16' 

D  = 

35 
"128 

Substituting  in  (1),  x  = 

2^^8^  ^16 

^  ^128 

y*  + 

,  etc. 


If  the  even  powers  of  x  are  wanting  in  the  given  series,  the 
operation  may  be  abridged  by  assuming  x  equal  to  a  series  con- 
taining only  the  odd  powers  of  y. 


EXERCISE  175 

Revert  each  of  the  following  to  four  terms : 


1.  y  =  aj+-3ar'+-5aj3-f7a;*+....        ^  .  x^  .  a^ 

2.  y  =  x-2x^-\-3a^-4.x'-^"'. 

4.   y  =  2x  +  5x^-{-Sa^  +  llx^  + 


3.  ,=.+_+_+_+.... 


^     2     4      6      8 


6.  3,=£+^+^V-+ 
*    [2    [3^[4^[5^ 


7.  2/  =  2a;-4a^  +  6a^-8a!'  + 
"     2468 


372  ALGEBKA 

XXX.    THE  BINOMIAL  THEOREM 

FRACTIONAL  AND  NEGATIVE  EXPONENTS 

407.  It  was  proved  in  §  387  that,  if  ti  is  a  positive  integer, 

(a  +  xy  =  a"  +  na--'x  +  ^  ^1"  ~  ^)  a"-^a^ 

1  •  2 

n(yi-l)  (71-2)^^.3^ 
1-2-3 

If  n  is  a  negative  integer,  or  a  positive  or  negative  fraction, 
the  series  in  tlie  second  member  is  infinite ;  for  no  one  of  the 
expressions  n  —  1,  n  —  2,  etc.,  can  equal  zero ;  in  this  case,  the 
series  gives  the  value  of  (a  -f-  xy,  provided  it  is  convergent. 

As  a  rigorous  proof  of  the  Binomial  Theorem  for  Fractional  and  Nega- 
tive Exponents  is  too  difficult  for  pupils  in  preparatory  schools,  the  author 
has  thought  best  to  omit  it ;  any  one  desiring  a  rigorous  algebraic  proof  of 
the  theorem,  will  find  it  in  the  author's  Advanced  Course  in  Algebra,  §  675. 

408.  Examples. 

In  expanding  expressions  by  the  Binomial  Theorem  when 
the  exponent  is  fractional  or  negative,  the  exponents  and 
coefficients  of  the  terms  may  be  found  by  the  laws  of  §  386, 
which  hold  for  all  values  of  the  exponent. 

1.   Expand  (a  +  xy  to  five  terms. 

2 
The  exponent  of  a  in  the  first  term  is  -,  and  decreases  by  1  in  each 

succeeding  term. 

The  exponent  of  x  in  the  second  term  is  1,  and  increases  by  1  in  each 
succeeding  term. 

The  coefficient  of  the  first  term  is  1  ;  of  the  second  term,  — 

2  i^ 

Multiplying  -,  the  coefficient  of  the  second  term,  by  — ,  the  exponent 

o  o 

of  a  in  that  term,  and  dividing  the  product  by  the  exponent  of  x  increased 

by  1,  or  2,  we  have  —  as  the  coefficient  of  the  third  term  ;  and  so  on. 
y 

Then,     (a  +  x)^  =  a^  +  -a~^  x  -  -a'^x"^  +  — a~^  x^  -  -^  a'^  x*  4-  —. 
'     ^         ^  3  9  81  '  243 


THE  BINOMIAL  THEOREM  373 

2.  Expand  (1  +  2  a;:^)-^  to  five  terms. 
Enclosing  2  x~^  in  parentheses,  we  have 
(1+2  x~^)-^  =  [1  +  (2  x~^)]-2 

=  1-2  _  2  .  1-3  .  (2  x~^)  +  3  .  1-*  .  (2  x'^y 

-  4  . 1-6  .  (2  «"^)3  +  5  . 1-6  .  (2  x~^y 

=  1  -  4  a;"^  +  12  x-^  -  32  x~^  +  80  x-2  +  .... 

By  writing  the  exponents  of  1,  in  expanding  [1  +  (2x~^)]-2,  we  can 
make  use  of  the  fifth  law  of  §  386. 

3.  Expand  —  to  four  terms. 

Enclosing  a"^  and  —  3x^  in  parentheses,  we  have 
^ ^ -[(a-i)+(-3a:br^ 


'^a-i-3x*     (a-i-3xbi 


=  («-^)'*  - 1  (a-0"^  (  -  3  xb  + 1  (a-^r^  (  -  3  x^^ 

-|f(a-^)-'^(-3xb«+... 

=  a^  +  a^x^  +  2  aW  +  ^ a'^a;  +  .... 
o 


EXERCISE  176 

Expand  each  of  the  following  to  five  terms : 

1.  (.  +  .)t.  6.  _^.  '''  ^C(a-65^cy]. 

^1-.  ,,^  1 


2.  (l  +  a.)-«.  7.  («f+2  6)l  (a.-^-22/V 

3.  (1  -  x)-K  8.  (a«  -  4  xY^.  13.  (^e!  +  ^'^-^. 

4.  ^^^76.  9.  ^_^^^J  14.  (^I_3n-^)-^. 


_  _ .     m^  +  — - 
(a  +  a^y  V  4 


5._A_.  10.  rm-  +  ^V^       15.'     ^         ---^ 


374  ALGEBRA 

409.  The  formula  for  the  rth  term  of  (a  -j-  xy  (§  390)  holds 
for  fractional  or  negative  values  of  n,  since  it  was  derived  from 
an  expansion  which  holds  for  all  values  of  the  exponent. 

Ex.  Find  the  7th  term  of  (a  -  3  x'^^yi 
Enclosing  —  3  x~^  in  parentheses,  we  have 

(a  -  3  x~^)~3  =  [a  ^.  (_  3  x"2)]~3. 

The  exponent  of  (-  3 x~^)  is  7  -  1,  or  6. 

1  19 

The  exponent  of  a  is 6,  or 

3  3 

1  IQ 

The  first  factor  of  the  numerator  is  — ,  and  the  last  factor  —  —  +  1, 

or--. 

The  last  factor  of  the  denominator  is  6. 
Hence,  the  7th  term 

_  1 .  _  §  .  _  Z  .  _  10  .  _  13  .  _  16 
3 '      3 '      3 '       3  *       3  '       3    _i_9  _3 

= 1.2.3.4.5.6 -«    ^  C-3^  ^)' 

38  ^         ^9 


EXERCISE  177 
Find  the : 


1.  6th  term  of  (a  4- a^)l         ^-    9th  term  of  (a  -  a;)-l 

2.  5th  term  of  (a  —  o)~i 

3.  7th  term  of  (1  +  x)-l 


^     ^  ,  „  ,  1  6.    11th  term  of  V(m  +  nY. 

2.    5th  term  of  (a  -  o)"^.  / 

7.    7th  term  of  (a'^  -  2  b^)-\ 


8.   8th  term  of 


4.   8th  term  of  (1  -  x)\  (a^  -f  y-iy 

9.    10th  term  of  (x-^  +  y^yK 
10.   6th  term  of  (a^  -  2  b-yK 
.    11.   5th  term  of  (m  +  3  n-^)i 

12.   9th  term  of 


THE  BINOMIAL  THEOREM  375 

13.  11th  term  of  ( a-W  -  — V^- 

14.  10th  term  of  {x~^  -  4  /)l 

410.   Extraction  of  Roots. 

The  Binomial  Theorem  may  sometimes  be  used  to  find  the 
approximate  root  of  a  number  which  is  not  a  perfect  power  of 
the  same  degree  as  the  index  of  the  root. 

Ex.   Find  ■\/25  approximately  to  five  places  of  decimals. 
The  nearest  perfect  cube  to  25  is  27. 


We  have  V25  =  V27  -  2  =  [(S^)  +  (-  2)] 


=  (33)i  + 1  (33)-f  (  -  2)  - 1  (33)-t (  -  2)2 

+  |^(3«)"^(-2)»-... 
=  3         ^  4  40 


3  .  32      9  .  35      81  •  38 

Expressing  each  fraction  approximately  to  the  nearest  fifth  decimal 
place,  we  have 

y/2b  =  3  -  .07407  -  .00183  -  .00008 =  2.92402. 

We  then  have  the  following  rule : 

Separate  the  given  number  into  two  parts,  the  first  of  which  is 
the  nearest  perfect  power  of  the  same  degree  as  the  required  root, 
and  expand  the  result  by  the  Binomial  Theorem. 

If  the  ratio  of  the  second  term  of  the  binomial  to  the  first  is  a  small 
proper  fraction,  the  terms  of  the  expansion  diminish  rapidly  ;  but  if  this 
ratio  is  but  little  less  than  1,  it  requires  a  great  many  terms  to  insure  any 
degree  of  accuracy. 

EXERCISE  178 

Find  the  approximate  values  of  the  following  to  five  places 
of  decimals : 

1.  Vl7.      2.  V51.      3.  ^/60.      4.  ^14.      5.  <M.     6.  ^35. 


376  ALGEBRA 


XXXI.    LOGARITHMS 

411.  The  Common  System. 

Every  positive  arithmetical  number  may  be  expressed,  exactly 
or  approximately,  as  a  power  of  10. 

Thus,  100  =  102;  13  =  10i"3»-;  etc. 

When  thus  expressed,  the  corresponding  exponent  is  called 
its  Logarithm  to  the  Base  10. 

Thus,  2  is  the  logarithm  of  100  to  the  base  10;  a  relation 
which  is  written  logio  1^^  =  2,  or  simply  log  100  =  2. 

Logarithms  of  numbers  to  the  base  10  are  called  Common 
Logarithms,  and,  collectively,  form  the  Common  System. 

They  are  the  only  ones  used  for  numerical  computations. 

412.  Any  positive  number,  except  unity,  may  be  taken  as 
the  base  of  a  system  of  logarithms ;  thus,  if  a'  =  m,  where  a 
and  m  are  positive  numbers,  then  x  =  log^  m. 

A  negative  number  is  not  considered  as  having  a  logarithm. 

413.  By  §§  238  and  239, 

10«  =  1,  10-^  =  1  =.1, 

10^  =  10,  10-2  =  ^^=.01, 

102  =  100,  10-3  =  r^  =  .001,  etc. 

Whence,  by  the  definition  of  §  411, 

log  1  =  0,  log  .1  =  -  1  =  9  - 10, 

log  10  =  1,  log  .01  =  -  2  =  8  - 10, 

log  100  =  2,  log  .001  =  -  3  =  7  - 10,  etc. 

The  second  form  for  log  .1,  log  .01,  etc.,  is  preferable  in  practice. 
If  no  base  is  expressed,  the  base  10  is  understood. 


LOGARITHMS  377 

414.  It  is  evident  from  §  413  that  the  common  logarithm  of 
a  number  greater  than  1  is  positive,  and  the  logarithm  of  a 
number  between  0  and  1  negative. 

415.  If  a  number  is  not  an  exact  power  of  10,  its  common 
logarithm  can  only  be  expressed  approximately ;  the  integral 
part  of  the  logarithm  is  called  the  characteristic,  and  the  decimal 
part  the  mantissa. 

For  example,  log  13  =  1.1139. 

Here,  the  characteristic  is  1,  and  the  mantissa  .1139. 
A  negative  logarithm  is  always  expressed  with  a  positive 
mantissa,  which  is  done  by  adding  and  subtracting  10. 

Thus,  the  negative  logarithm  —2.5863  is  written  7.4137  —  10. 
In  this  case,  7  — 10  is  the  characteristic. 

The  negative  logarithm  7.4137  —  10  is  sometimes  written  3.4137 ;  the 
negative  sign  over  the  characteristic  showing  that  it  alone  is  negative,  the 
mantissa  being  always  positive. 

For  reasons  which  will  appear,  only  the  mantissa  of  the 
logarithm  is  given  in  a  table  of  logarithms  of  numbers;  the 
characteristic  must  be  found  by  aid  of  the  rules  of  §§  416 
and  417. 

416.  It  is  evident  from  §  413  that  the  logarithm  of  a 
number  between 

1  and      10  is  equal  to  0  +  a  decimal ; 
10  and    100  is  equal  to  1  -f  a  decimal ; 
100  and  1000  is  equal  to  2  +  a  decimal ;  etc. 

Therefore,  the  characteristic  of  the  logarithm  of  a  number 
with  one  place  to  the  left  of  the  decimal  point  is  0 ;  with  two 
places  to  the  left  of  the  decimal  point  is  1 ;  with  three  places 
to  the  left  of  the  decimal  point  is  2 ;  etc. 

Hence,  the  characteristic  of  the  logarithm  of  a  number  greater 
than  1  is  1  less  than  the  number  of  places  to  the  left  of  the  decimal 
point. 

For  example,  the  characteristic  of  log  906328.51  is  5. 


378  ALGEBRA 

417.  In  like  manner,  the  logarithm  of  a  number  between 

1  and      .1  is  equal  to  9  -f-  a  decimal  —  10 ; 
.1  and    .01  is  equal  to  8  -f  a  decimal  —  10 ; 
.01  and  .001  is  equal  to  7  -f-  a  decimal  —  10 ;  etc. 

Therefore,  the  characteristic  of  the  logarithm  of  a  decimal 
with  no  ciphers  between  its  decimal  point  and  first  significant 
figure  is  9,  with  — 10  after  the  mantissa ;  of  a  decimal  with 
one  cipher  between  its  point  and  first  significant  figure  is  8, 
with  —10  after  the  mantissa;  of  a  decimal  with  two  ciphers 
between  its  point  and  first  significant  figure  is  7,  with  —10 
after  the  mantissa ;  etc. 

Hence,  to  find  the  characteristic  of  the  logarithm  of  a  number 
less  than  1,  subtract  the  number  of  ciphers  between  the  decimal 
point  and  first  significant  figure  from  9,  writing  — 10  after  the 
mantissa. 

For  example,  the  characteristic  of  log  .007023  is  7,  with  —10 
written  after  the  mantissa. 

PROPERTIES   OF  LOGARITHMS 

418.  In  any  system,  the  logarithm  of  1  is  0. 

For  by  §  238,  a*^  =  1 ;  whence,  by  §  412,  log«  1  =  0. 

419.  In  any  system,  the  logarithm  of  the  base  is  1. 
For,  a^  =  a ;  whence,  log^  a  =  1. 

420.  In  any  system  ijuhose  base  is  greater  than  1,  the  loganthm 
of  0  is  —oo. 

For  if  a  is  greater  than  1,  a"*  =  —  =  1  =  0  (§  320). 

a       oo 

Whence,  by  §  412,  logaO  =  -  oo. 

No  literal  meaning  can  be  attached  to  such  a  result  as  loga  0  =  —  oo  ; 
it  must  be  interpreted  as  follows  : 

If,  in  any  system  whose  base  is  greater  than  unity,  a  number  approaches 
the  limit  0,  its  logarithm  is  negative,  and  increases  indefinitely  in  absolute 
value.     (Compare  §  321.) 


LOGARITHMS  379 

421.   In  any  system^  the  logarithm  of  a  product  is  equal  to 
the  sum  of  the  logarithms  of  its  factors. 

Assume  the  equations 


a'  =  m 

a^  =  n 


whence,  by  §  412,  [  ^  "  J°^«  '^^ 
2/  =  logaW. 


Multiplying  the  assumed  equations, 

a"  y^a^  =  mn,  or  a''+^  =  mn. 
Whence,        log^  mn  =  x-^y=  log„  m  +  loga  n. 

In  like  manner,  the  theorem  may  be  proved  for  the  product 
of  three  or  more  factors. 

By  aid  of  §  421,  the  logarithm  of  a  composite  number  may 
be  found  when  the  logarithms  of  its  factors  are  known. 

Ex.   Given  log  2  =  .3010,  and  log  3  =  .4771 ;  find  log  72. 
log  72  =  log  (2  X  2  X  2  X  3  X  3) 

=  log2  +  log2  +  log2  +  log3  +  log3 

=  3  X  log2  +  2  X  log3  =  .9030  +  .9542  =  1.8672. 

EXERCISE  179 
Given 

log 2  =  .3010,  log 3  =  .4771,  log 5  =  .6990,  log 7=  .8451,  find: 

1.  log  15.         4.    log  125.        7.   log  567.         10.    log  187'5. 

2.  log  98.         5.    log  315.        8.    log  1225.       11.    log  2646. 

3.  log  84.         6.    log  392.        9.    log  1372.       12.    log  24696. 

422.  In  any  system,  the  logarithm  of  a  fraction  is  equal  to 
the  logarithm  of  the  numerator  minus  the  logarithm  '  of  the 
denominator. 

Assume  the. equations 


a''  =  m\  ■  f£c  =  log„m, 

I- ;    whence, 
ay  =  n  J 


|a;  =  log„m, 
l2/  =  log„n. 


380  ALGEBRA 


Dividing  the  assumed  equations, 

—  =  — ,  or  a*^=  -• 
a^      n  n 

Whence,  log„  —z=x  —  y  =  log^m  —  log„n. 


Ex.    Given  log  2  =  .3010 ;  find  log  5. 

log  6  =  log—  =  log  10  -  log  2  =  1  -  .3010  =  .6990. 
2 


EXERCISE  180 
Given  log  2  =  .3010,  log  3  =  .4771,  log  7  =  .8451,  find : 

1.  logY-.  4.    log  245.  7.    log  If.  10.    log  ^o^. 

2.  log  2^.  5.   log85f.  8.   log  375.  11.   log  46f 
^3.    logllj.          6.   log  175.          9.    log  If.  12.   log  2Jf 

423.  In  any  system,  the  logarithm  of  any  power  of  a  number 
is  equal  to  the  logarithm  of  the  number  multiplied  by  the  exponent 
of  the  power. 

Assume  the  equation  a*  =  m ;  whence,  x  =  log^  m. 

Raising  both  members  of  the  assumed  equation  to  the^th 

P         '        a^  —  m^ ;  whence,  log„  mP=px=p  log„  m. 

424.  In  any  system,  the  logarithm  of  any  root  of  a  number 
is  equal  to  the  logarithm  of  the  number  divided  by  the  index  of 
the  root. 

—  i  1 

For,  log,  Vm  =  log„(m'-)  =  -log„m  (§  423). 

425.  Examples. 

1.    Given  log  2  =  .3010 ;  find  log  2l 

log  2^  =  I  X  log  2  =  ^  X  .3010  =  .5017. 
3  3 

To  multiply  a  logarithm  by  a  fraction,  multiply  first  by  the  numerator, 
and  divide  the  result  by  the  denominator. 


LOGARITHMS  381 

2.  Given  log  3  =  .4771 ;  find  log  ^'3. 

Iog^^l2g3  =  illl  =  .0596. 
8  8 

3.  Given  log  2  =  .3010,  log  3  =  .4771,  find  log  (2^  x  3^). 
By  §  421,  log  (23-  x  3^)  =  log  2^  +  log  3? 

=  i  log  2  +  -  log  3  =  .1003  +  .5964  =  .0907. 

EXERCISE  181 
Given  log  2  =  .3010,  log  3  =  .4771,  log-7  =  .8451,  find : 


1.    log  2«. 

5.   log42«.          9. 

log  50'. 

13.    log  a/8. 

2.   log5^ 

6.    log  45^.       10. 

log  ^3. 

14.    log  a/54. 

3.   log3i 

7.   log63l       11. 

log  a/5. 

15.    log  a/225. 

4.   log  7l 

8.   log98i       12. 

log  ^/7. 

16.    log  a/162. 

17.  log  n                 21.    log  ^^. 

18.  log(|)i                                ^2 

19.  Iog(3^xl00i).     22^   ^^^2± 

20.  log  (5a/3).                          5' 

23. 

24. 

log^f- 

78 

log    ^'  . 

^^75 

426.    To  prove  the  relation 

log,m  =  J°^« 
log. 

Assume  the  equations 

:    whence, 

'ic  =  log„w 
. y  =  logjm 

^j 

From  the  assumed  equations,  a"  = 

X 

Taking  the 

yth  root  of  both  members,  oy  =  h 

•• 

Therefore, 

log„6  =  -,  or  2/  = 

y 

X 

log„6 

That  is. 

log.m  =  |^^' 
log 

382  ALGEBRA 

By  aid  of  this  relation,  if  the  logarithm  of  a  number  m  to 
a  certain  base  a  is  known,  its  logarithm  to  any  other  base  h 
may  be  found  by  dividing  by  the  logarithm  of  h  to  the  base  a. 

427.  To  prove  the  relation 

log6axlog„6  =  l. 
Putting  m  =  a  in  the  result  of  §  426, 

log.a  =  |M^  =  -l-(§419). 
log,  6      log,  6 

Whence,  logj  a  x  log,6  =  1. 

428.  In  the  common  system,  the  mantissce  of  the  logarithms  of 
numbers  having  the  same  sequence  of  figures  are  equal. 

Suppose,  for  example,  that  log  3.053  =  .4847. 

Then,  log  305.3  =  log(100  x  3.053)  =  log  100  +  log  3.053 

=  2  4-  .4847  =  2.4847 ; 
log  .03053  =  log  (.01  X  3.053)  =  log  .01  +  log  3.053 

=  8 -10  +  . 4847  =  8.4847- 10;  etc. 

It  is  evident  from  the  above  that,  if  a  number  be  multiplied 
or  divided  by  any  integral  power  of  10,  producing  another 
number  with  the  same  sequence  of  figures,  the  mantissas  of 
their  logarithms  will  be  equal. 

For  this  reason,  only  mantissse  are  given,  in  a  table  of  Com- 
mon Logarithms ;  for  to  find  the  logarithm  of  any  number,  we 
have  only  to  find  the  mantissa  corresponding  to  its  sequence  of 
figures,  and  then  prefix  the  characteristic  in  accordance  with 
the  rules  of  §§  416  and  417. 

This  property  of  logarithms  only  holds  for  the  common 
system,  and  constitutes  its  superiority  over  other  systems  for 
numerical  computation. 

429.  Ex.     Given  log2=.3010,  log3  =  .4771;  find  log .00432. 

We  have     log  432  =  log  (2*  x  3^)  =  4  log  2  +  3  log  3  =  2.0323. 


LOGARITHMS  383 

Then,  by  §  428,  the  mantissa  of  the  result  is  .6353. 
Whence,  by  §  417,  log  .00432  =  7.6353  -  10. 

EXERCISE  182 
Given  log  2  =  .3010,  log  3  =  .4771,  log  7  =  .8451,  find : 

1.  log  2.7.                   6.   log  .00000686.  11.  log  337.5. 

2.  log  14.7.                7.   log  .00125.  12.  log  3.888. 

3.  log  .56.                  8.   log  5^70.  13.  log  (4.5)«. 

4.  log  .0162.               9.   log  .0000588.  .    14.  log -s/SA. 

5.  log  22.5.               10.   log  .000864.  15.  log  (24.3)1 

USE  OF  THE  TABLE 

430.  The  table  (pages  384  and  385)  gives  the  mantissae  of 
the  logarithms  of  all  integers  from  100  to  1000,  calculated  to 
four  places  of  decimals. 

431.  To  find  the  logarithm  of  a  number  of  three  figures. 
Look  in  the  column  headed  "  Ko."  for  the  first  two  signifi- 
cant figures  of  the  given  number. 

Then  the  required  mantissa  will  be  found  in  the  corre- 
sponding horizontal  line,  in  the  vertical  column  headed  by 
the  third  figure  of  the  number. 

Finally,  prefix  the  characteristic  in  accordance  with  the 
rules  of  §§  416  and  417. 

For  example,       log  168  =  2.2253 ; 

log  .344  =  9.5366  - 10 ;  etc. 

For  a  number  consisting  of  one  or  two  significant  figures,  the 
column  headed  0  may  be  used. 

Thus,  let  it  be  required  to  find  log  83  and  log  9. 

By  §  428,  log  83  has  the  same  mantissa  as  log  830,  and  log  9 
the  same  mantissa  as  log  900. 

Hence,  log  83  =  1.9191,  and  log  9  =  0.9542. 


384 


ALGEBRA 


No. 

0 

1 

2 

3 

4 

6 

6 

7 

8 

9 

lO 

0000 

0043 

0086 

0128 

0170 

0212 

0253 

0294 

0334 

0374 

II 

0414 

0453 

0492 

0531 

0569 

0607 

0645 

0682 

0719 

0755 

12 

0792 

0828 

0864 

0899 

©934^- 

-09^ 

1004 

1038 

1072 

1106 

13 

"39 

"73 

T206 

1239 

1271 

1303 

1335 

1367 

1399 

1430 

14 

1461 

1492 

1523 

1553 

1584 

1614 

1644 

1673 

1703 

1732 

15 

1761 

1790 

1818 

1847 

1875 

1903 

1931 

1959 

1987 

2014 

16 

2041 

2068 

2095 

2122 

2148 

2175 

2201 

2227 

2253 

2279 

17 

2304 

2330 

2355 

2380 

2405 

2430 

2455 

2480 

2504 

2529 

18 

2553 

2577 

2601 

2625 

2648 

2672 

2695 

2718 

2742 

2765 

19 

2788 

2810 

2833 

2856 

2878 

2900 

2923 

2945 

2967 

2989 

20 

3010 

3032 

3054 

3075 

3096 

3118 

3139 

3160 

3181 

3201 

21 

3222 

3243 

3263 

3284 

3304 

3324 

3345 

3365 

3385 

3404 

22 

3424 

3444 

3464 

3483 

3502 

3522 

3541 

3560 

3579 

3598 

23 

3617 

3636 

3655 

3674 

3692 

37" 

3729 

3747 

3766 

3784 

24 

3802 

3820 

3838 

3856 

3874 

3892 

3909 

3927 

3945 

3962 

25 

3979 

3997 

4014 

4031 

4048 

4065 

4082 

4099 

4116 

4133 

26 

4150 

4166 

4183 

4200 

4216 

4232 

4249 

4265 

4281 

4298 

27 

4314 

4330 

4346 

4362 

4378 

4393 

4409 

4425 

4440 

4456 

28 

4472 

4487 

4502 

4518 

4533 

4548 

4564 

4579 

4594 

4609 

29 

4624 

4639 

4654 

4669 

4683 

4698 

4713 

4728 

4742 

4757 

30 

4771 

4786 

4800 

4814 

4829 

4843 

4857 

4871 

4886 

4900 

31 

4914 

4928 

4942 

4955 

4969 

4983 

4997 

501 1 

5024 

5038 

32 

5051 

5065 

5079 

5092 

5105 

5"9 

5132 

5145 

5159 

5172 

33 

5185 

5198 

5211 

5224 

5237 

5250 

5263 

5276 

5289 

5302 

34 

5315 

5328 

5340 

5353 

5366 

5378 

5391 

5403 

5416 

5428 

35 

5441 

5453 

5465 

5478 

5490 

5502 

5514 

5527 

5539 

5551 

36 

5563 

5575 

5587 

5599 

5611 

5623 

5635 

5647 

5658 

37 

5682 

5694 

5705 

5717 

5729 

5740 

5752 

5763 

5775 

5786 

38 

5798 

5809 

5821 

5832 

5843 

5855 

5877 

5899 

39 

59" 

5922 

5933 

5944 

5955 

5966 

5977 

5988 

5999 

6010 

40 

6021 

6031 

6042 

6053 

6064 

6075 

6085 

6096 

6107 

6117 

41 

6128 

6138 

6149 

6160 

6170 

6180 

6191 

6201 

6212 

6222 

42 

6232 

6243 

6253 

6263 

6274 

6284 

6294 

6304 

6314 

6325 

43 

6335 

6345 

6355 

6365 

6375 

6385 

6395 

6405 

6415 

6425 

44 

6435 

6444 

6454 

6464 

6474 

6484 

6493 

6503 

6513 

6522 

45 

6532 

6542 

6551 

6561 

6571 

6580 

6590 

6599 

6609 

6618 

46 

6628 

6637 

6646 

6656 

6665 

6675 

6684 

6693 

6702 

6712 

47 

6721 

6730 

6739 

6749 

6758 

6767 

6776 

6785 

6794 

6803 

48 

6812 

6821 

6830 

6839 

6848 

6857 

6866 

6875 

6884 

6893 

49 

6902 

691 1 

6920 

6928 

6937 

6946 

6955 

6964 

6972 

6981 

50 

6990 

6998 

7007 

7016 

7024 

7033 

7042 

7050 

7059 

7067 

51 

7076 

7084 

7093 

7101 

7110 

7118 

7126 

7135 

7143 

7152 

52 

7160 

7168 

7177 

7185 

7193 

7202 

7210 

7218 

7226 

7235 

53 

7243 

7251 

7259 

7267 

7275 

7284 

7292 

7300 

7308 

7316 

54 

7324 

7332 

7340 

7348 

7356 

7364 

7372 

7380 

7388 

7396 

Uo. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

LOGARITHMS 


385 


No. 

0 

1 

2 

3 

4 

6 

6 

7 

8 

9 

55 

7404 

7412 

7419 

7427 

7435 

7443 

7451 

7459 

7466 

7474 

56 

7482 

7490 

7497 

7505 

7513 

7520 

7528 

7536 

7543 

7551 

57 

7559 

7566 

7574 

7582 

7589 

7597 

7604 

7612 

7619 

7627 

58 

7634 

7642 

7649 

7657 

7664 

7672 

7679 

7686 

7694 

7701 

59 

7709 

7716 

7723 

7731 

7738 

7745 

7752 

7760 

7767 

7774 

60 

7782 

7789 

7796 

7803 

7810 

7818 

7825 

7832 

7839 

7846 

6x 

7853 

7860 

7868 

7875 

7882 

7889 

7896 

7903 

7910 

7917 

62 

7924 

7931 

7938 

7945 

7952 

7959 

7966 

7973 

7980 

7987 

63 

7993 

8000 

8007 

8014 

8021 

8028 

8035 

8041 

8048 

8055 

64 

8062 

8069 

8075 

8082 

8089 

8096 

8102 

8109 

8116 

8122 

65 

8129 

8136 

8142 

8149 

8156 

8162 

8169 

8176 

8182 

8189 

66 

8195 

8202 

8209 

8215 

8222 

8228 

8235 

8241 

8248 

8254 

67 

8261 

8267 

8274 

8280 

8287 

8293 

8299 

8306 

8312 

8319 

68 

8325 

8331 

8338 

8344 

8351 

8357 

8363 

S370 

8376 

8382 

69 

8388 

8395 

8401 

8407 

8414 

8420 

8426 

8432 

8439 

8445 

70 

8451 

8457 

8463 

8470 

8476 

8482 

8488 

8494 

8500 

8506 

71 

8513 

8519 

8525 

8531 

8537 

8543 

8549 

8555 

8561 

8567 

72 

8573 

8579 

8585 

8591 

8597 

8603 

8609 

8615 

8621 

8627 

73 

8633 

8639 

8645 

8651 

8657 

8663 

8669 

8675 

8681 

8686 

74 

8692 

8698 

8704 

8710 

8716 

8722 

8727 

8733 

8739 

8745 

75 

8751 

8756 

8762 

8768 

8774 

8779 

8785 

8791 

8797 

8802 

76 

8808 

8814 

8820 

8825 

8831 

8837 

8842 

8848 

8854 

8859 

77 

8865 

8871 

8876 

8882 

8887 

8893 

8899 

8904 

8910 

8915 

78 

8921 

8927 

f932 

8938 

8943 

8949 

8954 

8960 

8965 

8971 

79 

8976 

8982 

8987 

8993 

8998 

9004 

9009 

9015 

9020 

9025 

80 

9031 

9036 

9042 

9047 

9053 

9058 

9063 

9069 

9074 

9079 

81 

9085 

9090 

9096 

9101 

9106 

9112 

9117 

9122 

9128 

9133 

82 

9138 

9143 

9149 

9154 

9159 

9165 

9170 

9175 

9180 

9186 

83 

9191 

9196 

9201 

9206 

9212 

9217 

9222 

9227 

9232 

9238 

84 

9243 

9248 

9253 

9258 

9263 

9269 

9274 

9279 

9284 

9289 

85 

9294 

9299 

9304 

9309 

9315 

9320 

9325 

9330 

9335 

9340 

86 

9345 

9350 

9355 

9360 

9365 

9370 

9375 

9380 

9385 

9390 

87 

9395 

9400 

9405 

9410 

9415 

9420 

9425 

9430 

9435 

9440 

88 

9445 

9450 

9455 

9460 

9465 

9469 

9474 

9479 

9484 

9489 

89 

9494 

9499 

9504 

9509 

9513 

9518 

9523 

9528 

9533 

9538 

90 

9542 

9547 

9552 

9557 

9562 

9566 

9571 

9576 

9581 

9586 

91 

9590 

9595 

9600 

9605 

9609 

9614 

9619 

9624 

9628 

^^.P 

92 

9638 

9643 

9647 

9652 

9657 

9661 

9666 

9671 

9675 

9680 

93 

9685 

9689 

9694 

9699 

9703 

9708 

9713 

9717 

9722 

9727 

94 

9731 

9736 

9741 

9745 

9750 

9754 

9759 

9763 

9768 

9773 

95 

9777 

9782 

9786 

9791 

9795 

9800 

9805 

9809 

9814 

9818 

96 

9823 

9827 

9832 

9836 

9841 

9845 

9850 

9854 

9859 

9863 

97 

9868 

9872 

9877 

9881 

9886 

9890 

9894 

9899 

9903 

9908 

98 

9912 

9917 

9921 

9926 

9930 

9934 

9939 

9943 

9948 

9952 

99 

9956 

9961 

9965 

9969 

9974 

9978 

9983 

9987 

9991 

9996 
9 

No. 

0 

1 

2 

3 

4 

5 

6 

7 

~8~ 

386  ALGEBRA 

432.  To  find  the  logarithm  of  a  number  of  more  than  three 
figures. 

1.  Eequired  the  logarithm  of  327.6. 
We  find  from  the  table,        log  327  =  2.5145, 

log328  =  2.5159. 

That  is,  an  increase  of  one  unit  in  the  number  produces  an  increase  of 
.0014  in  the  logarithm. 

Then  an  increase  of  .6  of  a  unit  in  the  nuflftber  will  increase  the 
logarithm  by  .6  x  .0014,  or  .0008  to  the  neareaflourth  decimal  place. 

Whence,  log  327.6  =  2.5145  +  .0008  =  2.5153. 

In  finding  the  logarithm  of  a  number,  the  difference  between  the  next 
less  and  next  greater  mantissas  is  called  the  tabular  dij0%Yence ;  thus,  in 
Ex.  1,  the  tabular  difference  is  .0014. 

The  subtraction  may  be  performed  mentally. 

The  following  rule  is  derived  from  the  above :  J 

Find  from  the  table  the  ma7itissa  of  the  first  three  significant 
figures,  and  the  tabular  difference. 

Midtiply  the  latter  by  the  remaining  figures  of  the  number,  with 
a  decimal  point  before  them. 

Add  the  result  to  the  maritissa  of  the  first  three  figures,  and 
prefix  the  proper  characteristic. 

In  finding  the  correction  to  the  nearest  units'  figure,  the  decimal  por- 
tion should  be  omitted,  provided  that  if  it  is  .6,  or  greater  than  .5,  the 
units'  figure  is  increased  by  1 ;  thus,  13.26  would  be  taken  as  13,  30.5  as 
31,  and  22.803  as  23, 

2.  Find  the  logarithm  of  .021508.  ^ 

Mantissa  215  =  .3324  Tab.  diff.  =     21 

2  _M^ 

.3326  Correction  =  1.68  =  2,  nearly. 

The  result  is  8.3326  -  10. 

EXERCISE  183 
Find  the  logarithms  of  the  following: 
1.    64.  2.    3.7  3.    982.  4.    .798. 


LOGARITHMS  387 

5.  1079.  9.  .00005023.  13.  7.3165. 

6.  .6757.  10.  .0002625.  14.  .019608. 

7.  .09496.  -  11.  31.393.  15.  810.39. 

8.  4.288.  12.  48387.  16.  .0025446. 

433.    To  find  the  number  corresponding  to  a  logarithm. 
1.    Required  the  number  whose  logarithm  is  1.6571. 

Find  in  the  table  the  mantissa  6571. 

In  the  corresponding  line,  in  the  column  headed  "  No.,"  we  find  45,  the 
first  two  figures  of  the  required  number,  and  at  the  head  of  the  column  we       / 
find  4,  the  third  figure.  \^  * 

Since  the  characteristic  is  1,  there  must  be  two  places  to  the  left  -^^  *^f^  ^^ 

decimal  point  (§  416). 


of  the    ^'V 


Hence,  the  number  corresponding  to  1.6571  is  45.4. 

2.    Required  the  number  who^  Ipgarithm  is  2.3934. 

We  find  in  the  table  the  mantis&'3927  and  39^5.'      ■"  a 

The  numbers  corresponding  to  the  logarithms  2.3927  and  2.3945  are  a'TK 

247  and  248,  respectively.  "^  \    y 

That  is,  an  increase  of  .0018  in  the  mantissa  produces  an  increase  of  "^^/^ 

one  unit  in  the  number  corresponding.  r^O  ^ 

Then,  an  increase  of  .0007  in  the  mantissa  will  increase  the  number  by  \       ^ 

—  of  a  unit,  or  .4,  nearly.  dPy^ 

Hence,  the  number  corresponding  is  247  +  .4,  or  247.4.  « -<^, 

The  following  rule  is  derived  from  the  above :  A 

Fi7id  from  the  table  the  next  less  mantissa,  the  three  figures   j/ 
corresponding,  and  the  tabular  difference.  ^^ 

Subtract  the  next  less  from  the  given  mantissa,  and  divide  the      -i 
remainder  by  the  tabular  difference.  ''i,^ 

Ayinex  the  quotient  to  the  first  three  figures  of  the  number,  and 
point  off  the  result. 

The  rules  for  pointing  off  are  the  reverse  of  those  of  §§  416  and  417  : 

I.  ijf—  10  is  not  written  after  the  mantissa^  add  1  to  the  characteristic, 
giving  the  number  of  places  to  the  left  of  the  decimal  point. 

II.  If  —  \Q  is  written  after  the  mantissa,  subtract  the  ptositive  part  of 
the  characteristic  from  9,  giving  the  number  of  ciphers  to  be  placed  between 
the  decimal  point  and  first  significant  figure. 


388  ALGEBRA 

3.   Find  tKe  number  whose  logarithm  is  8.5265  — 10. 

5265 
Next  less  mant.  =  5263. ;  figures  corresponding,  336. 
Tab.  diff.  13)2.00(.15  =  .2,  nearly. 
13 
70 

By  the  above  rule,  there  will  be  one  cipher  to  be  placed  between  the 
decimal  point  and  first  significant  figure  ;  the  result  is  .03362. 

The  correction  can  usually  be  depended  upon  to  only  one  4ecimal 
place  ;  the  division  should  be  carried  to  two  places  to  determine  the  last 
figure  accurately. 

EXERCISE  184 

Find  the  numbers  corresponding  to  the  following : 

1.  0.8189.  -6.   8.7954-10.  11.   1.3019. 

2.  7.6064-10.         7.    6.5993-10.  12.   4.2527-10. 

3.  1.8767.  8.   9.9437-10.  13.   2.0159. 

4.  2.6760.  9.   0.7781.  14.  3.7264-10. 

5.  3.9826.  10.   5.4571-10.  15.   4.4929. 


APPLICATIONS 

434.  The  approximate  value  of  a  number  in  which  the 
operations  indicated  involve  only  multiplication,  division,  invo- 
lution, or  evolution  may  be  conveniently  found  by  logarithms. 

The  utility  of  the  process  consists  in  the  fact  that  addition 
takes  the  place  of  multiplication,  subtraction  of  division, 
multiplication  of  involution,  and  division  of  evolution. 

1.   Find  the  value  of  .0631  x  7.208  x  .51272. 
By  §  421,  log  (.0631  x  7.208  x  .51272) 

=  log  .0631  +  log  7.208  +  log  .51272. 

log   .0631=   8.8000-10 
'^      -log   7.208=   0.8578 

log  .51272=   9.7099-10 
Adding,  log  of  result  =  19.3677  -  20  =  9.3677  -  10  (See  Note  1.) 

Number  corresponding  to  9.3677  -  10  =  .2332. 


LOGARITHMS  389 

Note  1.  If  the  sum  is  a  negative  logarithm,  it  should  be  written  in 
such  a  form  that  the  negative  portion  of  the  characteristic  may  be  —  10. 

Thus,  19.3677  -  20  is  written  9.3677  -  10. 

(In  computations  with  four-place  logarithms,  the  result  cannot  usually 
be  depended  upon  to  more  than  four  significant  figures.) 

"  336.8 


2.   Find  the  value  of 


7984 


By  .§  422,  log  ^  =  log  336.8  -  log  7984. 

log  336.8  =  12.5273 -10 
log  7984  =   3.9022 
Subtracting,  log  of  result  =   8.6251  -  10  (See  Note  2.) 

Number  corresponding  =  .04218. 

Note  2.  To  subtract  a  greater  logarithm  from  a  less,  or  a  negative 
logarithm  from  a  positive,  increase  the  characteristic  of  the  minuend  by 
10,  writing  —  10  after  the  mantissa  to  compensate. 

Thus,  to  subtract  3.9022  from  2.5273,  write  the  minuend  in  the  form 
12.5273  -  10  ;  subtracting  3.9022  from  this,  the  result  is  8.6251  -  10. 

3.   Find  the  value  of  (.07396)^ 
By  §  423,  log  (.07396)5  =    5  x  log  .07396. 

log  .07396  =    8.8690  -  10 


44.3450-50 
=    4.3450  -  10  =  log  .000002213. 


4.   Find  the  value  of  V.035063. 

By  §  424,  log  v^.035063  =  i  log  .035063. 

o 

log  .035063  =  8.5449- 10 

3)28.5449-30     (See  Note  3.) 
9.5150 -10  =  log  .3274. 

Note  3.  To  divide  a  negative  logarithm,  write  it  in  such  a  form  that 
the  negative  portion  of  the  characteristic  may  be  exactly  divisible  by  the 
divisor,  with  —  10  as  the  quotient. 

Thus,  to  divide  8.5449  —  10  by  3,  we  write  the  logarithm  in  the  form 
28.5449  -  30  ;  dividing  this  by  3,  the  quotient  is  9.5150  -  10. 


390 


ALGEBRA 


EXERCISE  185 

A  negative  number  has  no  common  logarithm  (§  412);  if  such  numbers 
occur  in  computation,  they  may  be  treated  as  if  they  were  positive,  and 
the  sign  of  the  result  determined  irrespective  of  the  logarithmic  work. 

Thus,  in  Ex.  3  of  the  following  set,  to  find  the  value  of  (-  95.86)  x  3.3918 
we  find  the  value  of  95.86  x  3.3918,  and  put  a  —  sign  before  the  result. 


Find  by  logarithms  the  values  of  the  following : 

1.  4.253x7.104.  4.    54.029  x  (- .0081487). 

2.  6823.2  X  .1634.         5.  .040764  x  .12896. 

3.  (-  95.86)  X  3.3918.      6.  (-285.46)  x  (-  .00070682). 


8. 


9. 


10. 


11, 


5978 
9.762' 

21.658 
45057  * 

.06405 
.002037' 

-38.19 
.10792  * 

670.43 
-^5382.3* 


12 


.000007913 


.00082375 

13.  (88.08)1 

14.  (.09437)^ 

15.  (3.625)^ 

16.  (-.4623)^ 

17.  loot 

18.  (.09)1 

19.  (85.7)^. 


20.  (-.000216)1 

21.  V7. 

22.  </3. 

23.  V"^. 

24.  VVd. 

25.  -v/:2005. 


26.  V.08367. 


27.  V.00015027. 

28.  \/'^^^:0040628. 


435.   Arithmetical  Complement. 

The  Arithmetical  Complement  of  the  logarithm  of  a  number, 
or,  briefly,  the  Cologarithm  of  the  number,  is  the  logarithm  of 
the  reciprocal  of  that  number. 

Thus,  colog  409  =  log  ^-^  =  log  1  -  log  409. 

log  1  =  10.         - 10     (See  Ex.  2,  §  434.) 
log  409  =   2.6117 
.-.  colog  409  =   7.3883-10. 

1 


Again,         colog  .067  =  log 


.067 


log  1  -  log  .067. 


LOGARITHMS  891 

log  1=10.         -10 
log  .067  =   8.8261-10 
.  •.  colog  .067  =   1.1739. 

It  follows  from  the  above  that  the  cologarithm  of  a  number 
may  be  found  by  subtracting  its  logarithm  from  10  — 10. 

The  cologarithm  may  be  found  by  subtracting  the  last  significant  figure 
of  the  logarithm  from  10  and  each  of  the  others  from  9,-10  being 
written  after  the  result  in  the  case  of  a  positive  logarithm. 

.51384 


Ex. 


log 


a.   oiio    V . 

^^  8.708  X  .0946 

.51384 
8.708  X  .0946 

.log (.51384  x^^^^^x_^y 

=  log.51384  +  log^}^^  +  log^^l^^ 

=  log  .51384  +  colog  8.708  +  colog  .0946. 

log. 

51.384 

=  9.7109-10 

colog 

8.708 

=  9.0601  -  10 

colog 

.0946 

=  1.0241 

9.7951  -  10  =  log  .6239. 

It  is  evident  from  the  above  example  that,  to  find  the  loga- 
rithm of  a  fraction  whose  terms  are  the  products  of  factors,  we 
add  together  the  logarithms  of  the  factors  of  the  numerator,  and 
the  cologarithms  of  the  factors  of  the  denominator. 

The  value  of  the  above  fraction  may  be  found  without  using  cologa- 
rithms, by  the  following  formula  : 

log i^l^§^ =  log  .51384  -  log  (8.709  x  .0946) 

^  8.709  X. 0946         ^  ^^  ^ 

=  log  .51384  -  (log  8.709  +  log  .0946). 

The  advantage  in  the  use  of  cologarithms  is  that  the  written  work  of 
computation  is  exhibited  in  a  more  compact  form. 


MISCELLANEOUS  EXAMPLES 

436.   1.   Find  the  value  of  ^^^ 

3^ 


392  ALGEBRA 


log  ^  =  log  2  +  log  v^5  +  colog  3^     (§  435) 

3^  1  ^ 

=  log2  +  Mog5  +  ^colog3. 
6  b 


log  2=    .3010 
log  5=    .6990;  -r- 3  =    .2330 

colog  3  ^  9.5229  -  10  ;   x  -  =  9.6024  -  10 

.1364       =logl. 


2.    Find  the  value  of  \l~'^^!^^^' 
V     7.962 


>'7.9 


03_296  ^  1  i,g  :03296  ^  1  ^^^^^  _        ^ 

962       3     ^  7.962       3  ^    ^  &  J 

log  .03296  =  8.5180  -  10 
log    7.962  =  0.9010 

3)27.6170-30 

9.2057 -10  =  log.  1606. 
The  result  is  -  .1606. 

EXERCISE   186 

Find  by  logarithms  the  values  of  the  following : 

J     2078.5  X  .05834  .  «     (-.076917)  x  26.3 

.3583x346    *  *    .5478  x  (-3120.7)* 


2. 


(-6.08)  X. 1304  .         .8102  x(- 6.225) 

4.046  X  .0031095*  '    (-  .0721)  x  (-  17.976)' 


5.    6^x5i  .f.     f     5510\^  14  ^4/7       8/3 

^    ^'  ,3gy  15.  V6X  ^10x^2. 

^      n/68  ^^'    VsoTg*  16  f ^^-^^     'V , 

^'    SIsS'  '  \     8.7  X. 06037 

g   _^^  ..12.  ^^^=f^.  jy  VTMMe 

(.1)1*  ^'  *  ^/.0003867 

9       (100)»   ,  j3      -(.03)^  18.  (--^4582)^. 

^  _  .004  -s/  - 1000  -  (.72346)^ 


LOGARITHMS  893 

19.  (- 143.59)"  x(.00532)^       ^^  (.0462)^ 

20.  ^40.954  X  .0002098.  '    ^^8.27  x  V:2296' 

i"  ^  OA     -^- 7.92  X  (.1807)^ 

21.  (3075.6)*  X  (.016432)^.         24.    ^^^^^^^ Z_. 

22     -v^28l8  X  -v^MO,  25  -27.931 

^61021  -\/M6  X  (.03023)7 

26.    -\/-  .067268  x  ^-.4175  x  a/.00263. 
27     .0005616  x  a/424:6  28     485.7  x  (.7301)^  x  -S^TOOO 

(6.73)^  X  (.03194)^  (9.127)«  x  (.7095)^ 

EXPONENTIAL  EQUATIONS 

437.  An  Exponential  Equation  is  an  equation  in  which  the 
unknown  number  occurs  as  an  exponent. 

To  solve  an  equation  of  this  form,  take  the  logarithms  of  both 
members ;  the  result  will  be  an  equation  which  can  be  solved  by 
ordinary  algebraic  methods. 

1.  Given  31^  =  23  ;  find  the  value  of  x. 
Taking  the  logarithms  of  both  members, 

log  (31^)  =  log  23  ;    or  a:  log  31  =  log  23  (§  423). 

Then,  ^^I2g23^0617^9,3 

log  31      1.4914 

2.  Solve  the  equation  .2^  =  3. 

Taking  the  logarithms  of  both  members,  x  log  .2  =  log  3. 

Then,      x  =  i^  =  — iII^^-i^  =  -.0285  +  . 
log. 2      9.3010-10      -.699 

An  equation  of  the  form  w"  =  h  may  be  solved  by  inspection 
if  b  can  be  expressed  as  an  exact  power  of  ^. 

3.  Solve  the  equation  16*  =  128. 

We  may  write  the  equation  (2*)=«  =  2^,  or  2^"  =  2^. 

7 
Then,  by  inspection,  Ax  =  l  \  and  x  =  — 

4 


394  ALGEBRA 

(If  the  equation  were  16*  =  -^,  we  could  write  it  (2*)*  =  —  =  2"'^ ; 
y^  128  2' 

then  4  x  would  equal  —  7,  and  x 


-I) 


EXERCISE  187 

Solve  the  following  equations  : 

1.  13^  =  8.  5.   34^-1  =  42^+1  9.   32-=gV 

2.  .06^  =  .9.  6.    7^-+2^.8^  10.    (tV)'  =  8. 

3.  9.347=^  =  . 0625.  7.    .2^+^  =  .5^-^  11.    (i)^=2V 

4.  .005038^  =  816.3.        8.    16^  =  32. 

12.  Given  a,  r,  and  l]  derive  the  formula  for  n  (§  372). 

13.  Given  a,  r,  and  S ;  derive  the  formula  for  n. 

14.  Given  a,  I,  and  S ;  derive  the  formula  for  n. 

15.  Given  ?*,  I,  and  S-,  derive  the  formula  for  n. 

438.   1.   Find  the  logarithm  of  .3  to  the  base  7. 

By  S  426,  log,  .3  =  Ip^  =  «:«Ii-^  =  ---^  =  -  ■^^^^-- 
logio7  .8451  .84ol 

Examples  of  this  kind  may  be  solved  by  inspection,  if  the 
number  can  be  expressed  as  an  exact  power  of  the  base. 

2.    Find  the  logarithm  of  128  to  the  base  16. 

Let  logic  128  =  x  ;  then,  by  §  412,  16^  =  128. 

Then,  as  in  Ex.  3,  §  437,  x  =  -;  that  is,  logie  128  =  I- 
4  4 


EXERCISE  188 
Find  the  values  of  the  following : 

1.  log7  59.  3.   log.482.  5.  log68  2.92. 

2.  log6.7.  4.   log.9 .00453.  6.   logsi  .0604. 

Find  by  inspection  the  values  of  the  following : 

7.   log«125.      8.   log«(|).      9.   log^;^(3).      10.   log^,^_(rh)- 


MISCELLANEOUS  TOPICS  395 


XXXII.      MISCELLANEOUS  TOPICS 

HIGHEST    COMMON    FACTOR    AND    LOWEST    COMMON 
MULTIPLE    BY  DIVISION 

439.  We  will  now  show  how  to  find  the  H.  C.  F.  of  two 
polynomials  which  cannot  be  readily  factored  by  inspection. 

The  rule  in  Arithmetic  for  the  H.  C.  F.  of  two  numbers  is  : 

Divide  the  greater  number  by  the  less. 

If  there  be  a  remainder,  divide  the  divisor  by  it;  and  con- 
tinue thus  to  make  the  remainder  the  divisor,  and  the  preceding 
divisor  the  dividend,  until  there  is  no  remainder. 

The  last  divisor  is  the  H.  C.  F.  required. 

Thus,  let  it  be  required  to  find  the  H.  C.  F.  of  169  and  546. 

169)546(3 
507 

39)169(4 
156 
13)39(3 
39 
Then,  13  is  the  H.  C.  F.  required. 

440.  We  will  now  prove  that  a  rule  similar  to  that  of  §  439 
holds  for  the  H.  C.  F.  of  two  algebraic  expressions. 

Let  A  and  B  be  two  polynomials,  arranged  according  to  the 
descending  powers  of  some  common  letter. 

Let  the  exponent  of  this  letter  in  the  first  term  of  A  be 
equal  to,  or  greater  than,  its  exponent  in  the  first  term  of  B. 

Suppose  that  B  is  contained  \n  A  p  times,  with  a  remainder 
(7;  that  C  is  contained  in  B  q  times,  with  a  remainder  D\  and 
that  D  is  contained  in  O  r  times,  with  no  remainder. 

To  prove  that  D  is  the  H.  C.  F.  of  A  and  B. 

The  operation  of  division  is  shown  as  follows. 


396  ALGEBRA 

B)A{p 
pB 

~~C)B(q 

D)C{r 
rD 

We  will  first  prove  that  Z>  is  a  common  factor  of  A  and  B. 
Since   the   minuend   is   equal  to  the  subtrahend   plus  the 
remainder  (§  34), 

A=pB  +  G,  (1) 

B=qC  +  D,  (2) 

and  C  =  rD. 

Substituting  the  value  of  C  in  (2),  we  obtain 

B=qrD^-D  =  D{qr  +  l).  (3) 

Substituting  the  values  of  B  and  C  in  (1),  we  have 

A=pD{qr-\-l)  +  rD  =  D{pqr+p  +  r).  (4) 

From  (3)  and  (4),  i)  is  a  common  factor  of  A  and  B. 
We  will  next  prove  that  every  common  factor  of  A  and  B  is 
a  factor  of  D. 

Let  F  be  any  common  factor  of  A  and  B ;  and  let 

A  =  mF,  and  B  =  nF. 
From  the  operation  of  division,  we  have 

C=A-pB,  (5) 

and  D  =  B-qC.  (6) 

Substituting  the  values  of  A  and  B  in  (5),  we  have 

(7=  mF  —  priF. 
Substituting  the  values  of  B  and  C  in  (6),  we  have 
D  =  nF  —  q{mF  —  pnF)  =  F(n  —  qm  +  2^9'^)- 


MISCELLANEOUS  TOPICS  397 

Whence,  i^is  a  factor  oi  D.  > 

Then,  since  every  common  factor  of  A  and  i9  is  a  factor  of 
D,  and  since  D  is  itself  a  common  factor  of  A  and  B,  it  follows 
that  D  is  the  highest  common  factor  of  A  and  B. 

We  then  have  the  following  rule  for  the  H.  C.  F.  of  two 
polynomials,  A  and  B,  arranged  according  to  the  descending 
powers  of  some  common  letter,  the  exponent  of  that  letter  in 
the  first  term  of  A  being  equal  to,  or  greater  than,  its  exponent 
in  the  first  term  of  B : 

Divide  A  by  B. 

If  there  be  a  remainder,  divide  the  divisor  by  it;  and  continue 
thus  to  make  the  remainder  the  divisor,  and  the  preceding  divisor 
the  dividend,  until  there  is  no  remainder. 

The  last  divisor  is  the  H.  C.  F.  required. 

It  is  important  to  keep  the  work  throughout  in  descending  powers  of 
some  common  letter ;  and  each  division  sliould  be  continued  until  the 
exponent  of  this  letter  in  the  first  term  of  the  remainder  is  less  than  its 
exponent  in  the  first  term  of  the  divisor. 

Note  1.  If  the  terms  of  one  of  the  expressions  have  a  common  factor 
which  is  not  a  common  factor  of  the  terms  of  the  other,  it  may  be  re- 
moved ;  for  it  can  evidently  form  no  part  of  the  highest  common  factor. 

In  like  manner,  we  may  divide  any  remainder  by  a  factor  which  is  not 
a  factor  of  the  preceding  divisor. 

1.   Find  the  H.  C.  F.  of 

6  a^- 25  a; +  14  and  6  x^  -  7  x"  -  25  x  +  IS. 

6ic2-25x+14)6a;3-    7  x'^ -26x  +  lS(x  +  S 
6  a;3  -  25  x2  +  14  a; 
18  x2  -  39  X 
18  x2  -  75  a;  +  42 
36  X  -  24 

In  accordance  with  Note  1,  we  divide  this  remainder  by  12,  giving 

*~    *  3x-2)6x2-25x  +  14(2ic-7 

6x2-  ix 
-21x 
-  21  X  +  14 

Then,  3  x  -  2  is  the  H.  C.  F.  required. 


398  ALGEBRA 

Note  2.  If  the  first  term  of  the  dividend,  or  of  any  remainder,  is  not 
divisible  by  the  first  term  of  the  divisor,  it  may  be  made  so  by  multiply- 
ing the  dividend  or  remainder  by  any  term  which  is  not  a  factor  of  the 
divisor. 

2.   Find  the  H.  C.  F.  of 

3a^-\-d'b-2  ab'  and  Aa^b  +  2  w'b^  -  ab^  +  b\ 

We  remove  the  factor  a  from  the  first  expression  and  the  factor  b  from 
the  second  (Note  1),  and  find  the  H.  C.  F.  of 

Sa^  +  ab-2b^  and  4a^-\-2  a^b  -  ab'^  +  b^ 

Since  4  a^  is  not  divisible  by  3  a'^,  we  multiply  the  second  expression  by 
3  (Note  2). 

4  a3  +  2  a^b  -     ab^  +    b'^ 


3  a2  +  a6  -  2  b-)\2  d^  +  %a%-2>  ab'-  +  3  Z>3(4  a 
12  gs  4, 4  a2j)  _  g  a&2 

2  a^b  +  5  a62  _^  3  ^3 

Since  2  a%  is  not  divisible  by  3  a^,  we  multiply  this  remainder  by 
3  (Note  2). 

2«26+   5a&2^    3&3 

3 

3  a2  +  a6  _  2  62)6  (^25  +  15  ^^2  +   9  63(2  ft 
Qa^b+   2ab^-   4  63 
13  a62  +  13  68 

We  divide  this  remainder  by  13  62  (Note  1),  giving  a  +  6. 

a  +  6)3  a2  +    a6  -  2  62(3  a  -  2  6 
Sa^  +  Sab 
-2ab 
-  2  a6  -  2  62 


Then,  a  +  6  is  the  H.  C.  F.  required. 

Note  3.  If  the  first  term  of  any  remainder  is  negative,  the  sign  of 
each  term  of  the  remainder  may  be  changed. 

Note  4.  If  the  given  expressions  have  a  common  factor  which  can 
be  seen  by  inspection,  remove  it,  and  find  the  H.  C.  F.  of  the  resulting 
expressions  ;  the  result,  multiplied  by  the  common  factor,  will  be  the 
H.  C.  F.  of  the  given  expressions. 


MISCELLANEOUS  TOPICS  399 

3.   Find  the  H.  C.  F.  of 

2x'-h3x'-6x--{-2x  and  6  a;H-  5  x^  -  2  o^  -  a;. 
Removing  the  common  factor  x  (Note  4),  we  find  the  H.  C.  F.  of 

2  x3  +  3 x2  -  6  a;  +  2  and  6x^+ 5x^ -2x-l. 

2a;3  +  3x2-6x  +  2)6x3  +  5ic2-   2x-l(3 
6  x3  +  9  ic2  -  18  X  +  6 
-  4  x2  +  16  X  -  7 

The  first  term  of  this  remainder  being  negative,  we  change  the  sign  of 
each  of  its  terms  (Note  3). 

2x3+3x2-6x4-2 
2 


4x2 -16x  + 7)4x3+    6x2-    12  x  +    4(x 
4x3-16x2+      7x 

22x2-    19  x+    4 
2 


44x2-    38x+    8(11 
44x2-176x  +  77 
69)138  X- 69 
2x-    1 
2x- 1)4x2-    16  x+    7(2x-7 
4  x2  —      2  X 

-  14x 

-  14x  +  7 


The  last  divisor  is  2  x  —  1  ;  multiplying  this  by  x,  the  H.  C.  F.  of  the 
given  expressions  is  x(2  x  —  1). 

(In  the  above  solution,  we  multiply  2  x3  +  3  x2  —  6  x  +  2  by  2  in  order 
to  make  its  first  term  divisible  by  4  x2  ;  and  we  multiply  the  remainder 
22  x2  —  19  X  +  4  by  2  to  make  its  first  term  divisible  by  4  x2.) 

EXERCISE  189 

Find  the  H.  C.  F.  of  the  following : 

1.  2a^  +  a-6,  Aa^-Sa-\-S. 

2.  Qx'-lTx-^-lO,  9a^-Ux-S. 

3.  x'-6x-27,  a^-2x'-Sx  +  21. 


400  ALGEBRA 

4.  Gx^-x-2,  Sx'-Ux'-x-^-e. 

5.  2'ia'-22ab-7b',  32a'-12ab-5b\ 

6.  16a^-hSx'y-\-13xy^  +  3f,  24.x^-Ux'y-\-13xy'-15  y^ 

7.  4:a^  +  4:X^-3x,  6x^ -\-llx^-x^-6x. 

8.  AxFy-Wxy^-j-dy^  S  x'  -IS  x^'y  +  25  x'y'  -12  xf. 

9.  6  a^  -{-5  a'  -6  a'  -3  a^  +  2  a",  9  a'  -i-lS  a'-^5  a'-  8  a  -4. 

10.  3  a^  - 13  a'b  +  3  a'b'-  +  4  a&^  9  a^b  + 12  a^^s  _  g  at^  _  5  b\ 

11.  4a^+  9aj  -9,  4x^  +  10ar''-7a;2  +  9. 

12.  6a*-7a2-5a2  +  5a-3,  S  a'-Qa'- 5  a' -9. 

13.  3  ^3  +  8  w^a;  -  9  nx-  +  2  ar^, 

6n' -^23  n^x-\-2  n^x" -13  nx^  +  2 x\ 

14.  a3  +  9a2  4-13a-15,  a^  +  9  (1^  +  22  fr'^  +  9  a2-9  a. 

15.  m«-27m^  m«4-4m^-25m^  +  12  ml 

16.  9a^  +  30a^&-21a2&2-f-12a&^ 

16  a^b  +  60  a^ft^  _  20  ab^  - 16  &^ 

17.  4:x'-llxy-20y%  2  x^  -  4.  x'y  -  17  x^  +  xy^  -\- 12  y". 

18.  4a^  +  8a4-15a^  +  2a2-4a,  Aa^-12a'-{-9a^-3a-\-2. 

19.  3a;3-8a^  +  16a;-8,  3x*-5a:3  +  5r^-lla;  +  6. 

20.  3a^^2_2a;y_7^^4_^7^2^5^3^^6^ 

3  a^2/'  +  7  a;y  +  5  ar^/-  5  aff-2xy^ 

21.  2a;*  +  5a^  +  4a^  +  7a;  +  6,  2  a;*  -  5  aj^  4.  n  a;2  _  9  3,  _|_  9^ 

22.  6a;*4-a^  +  3a^-6aj-4,  12  a;4  +  8  ar^-3  aj^-lO  a;-4. 

23.  3a^-8aj2-5a;  +  6,  a^- 5x^  +  5  a^  +  «2_^7  a;-3. 

441.   The  H.  C.  F.  of   three   expressions,  which  cannot  be 
readily  factored  by  inspection,  may  be  found  as  follows : 

Let  A,  B,  and  C  be  the  expressions. 

Let  G  be  the  H.  C.  F.  of  A  and  B ;  then,  every  common  factor 
of  G  and  C  is  a  common  factor  of  A,  B,  and  C. 


MISCELLANEOUS  TOPICS  401 

But  since  every  common  factor  of  two  expressions  exactly 
divides  their  H.  C.  F.,  every  common  factor  of  A,  B,  and  C  is 
also  a  common  factor  of  G  and  C. 

Whence,  the  H.  C.  F.  of  G  and  C  is  the  H.  C.  F.  of  A,  B, 
and  C. 

Hence,  to  find  the  H.  C.F.  of  three  expressions,  find  the  H.  C.F. 
of  two  of  them,,  and  then  of  this  result  and  the  third  expression. 

We  proceed  in  a  similar  manner  to  find  the  H.  C.  F.  of  any 
number  of  expressions. 

Ex.   Find  the  H.  C.  F.  of 

a;3_7a;  +  6,  ar^  +  3  a;^  -  16  ic  + 12,  and  »»- 5  a;2  +  7  a;-3. 

The  H.  C.  F.  of  x3  -  7  X  +  6  and  x3  +  3  x2  -  16  X  +  12  is  x^  -  3  x  +  2. 
The  H.  C.  F.  of  x-2  -  3  X  +  2  and  x3  -  5  x2  +  7  X  -  3  is  X  -  1. 

EXERCISE  190 

Find  the  H.  C.  F.  of  the  following : 

1.  6a;2-5a;-25,  9x^  +  27  a;  +  20,  12 ic^  + 11  a; - 15. 

2.  20a2H-23a5-7  52,  28^^-43  a6+9  6^   24.a'-\-U  ab-5b'. 

3.  5a2-33a-14,  5a^-13a--^Ua-{-S,  5aM-27a2+20a+4. 

4.  Sx'-Gxy-SBy^  10  x" -27  xhj -xy' -^15  f, 

6x^-13x^y-rS  xy-  +  20  y\ 

5.  a53^4a;2-llic  +  30,  ar*^  +  2  a;2-5  a;-6,  ar^  -  a;^  -  17  a;  -  15. 

6.  a3_8a2  +  20a-16,  a^  +  3a''-4.a-12, 

a3_6a2  +  lla-6. 

7.  3a3  +  17a26  +  18a62_85^  6a-^  4-a'6- 19  a6-  + 6  6^ 

8.  3a^-a;2_38^_24^  3^:3^5  ^_53  ^_40^ 

3a:3  +  26a;2_^61a.'  +  30. 

442.  We  will  now  show  how  to  find  the  L.C.M.  of  two 
expressions  which  cannot  be  readily  factored  by  inspection. 


402  ALGEBRA 

Let  A  and  B  be  any  two  expressions. 

Let  F  be  their  H.  C.  F.,  and  M  their  L.  C.  M. 

Suppose  that  A  =  aF,  and  B  =  bF. 

Then,  AxB  =  abF\  (1) 

Since  F  is  the  H.  C.  F.  of  A  and  jB,  a  and  h  have  no  common 
factors  5  whence,  the  L.  C.  M.  of  aF  and  hF  is  a6i^. 

That  is,  M=abF. 

Multiplying  each  of  these  equals  by  F,  we  have 

FxM=ahF\  (2) 

From  (1)  and  (2),  AxB  =  FxM.  (Ax.  4,  §  9) 

That  is,  the  product  of  two  expressions  is  equal  to  the  product 

of  their  H.  C.  F.  and  L.  C.  M. 

Therefore,  to  find  the  L.  C.  M.  of  two  expressions. 
Divide  their  product  by  their  highest  common  factor ;  or, 
Divide  one  of  the  expressio7is  by  their  highest  common  factor ^ 

and  multiply  the  quotient  by  the  other  expression. 

Ex.   Find  the  L.  C.  M.  of 

6a:2-17x  +  12  and  12x2-4i»-2L 

6 a;2 _  17  x+ 12)12x2-    4x-21(2 
12  x2  -  34  a;  +  24 


15)30  X  -  45 

2x-   3)r)x2-17x  +  12(3a;- 
6x2-    9x 

-4 

-  8x 

-  8X  +  12 

Then,  the  H.  C.  F.  of  the  expressions  is  2  x  —  3. 

Dividing  6  x^  —  17  x  +  1 2  by  2  x  —  3,  the  quotient  is  3  x  —  4. 

Then,  the  L.  C.  M.  is  (3  x -4) (12x2 -4  x -21). 

EXERCISE  191 

Find  the  L.  C.  M.  of  the  following : 

1.   Sx'-{-Ux-24,  3i»2_|_23aj  +  30. 


MISCELLANEOUS  TOPICS  403 

2.  6x'-31x7j  +  lSy%  9  a^ -hl5xy-Uf. 

3.  4aj2  +  13a;  +  3,  4a^-23a;-6. 

4.  Sx'  +  6x-9,  ex^  +  Tx'-Tx-e. 

5.  3  a^  -  8  a'b  +  4  a^^^  a^6  -  11  a'b^  -f  22  aft^  -  8  b\ 

6.  6w3+n^i«-llw^'-6a^,  6  7i3-5  7i2iz;-8na^2_p3a^. 

7.  2a;^  +  7a^H-7a^2_|_2a;^  2  a;^  +  a;' - 10  a;^  -  8  a;. 

8.  6a^  +  aj2-17ic4-10,  3aj^  +  5ar^-5a;2_5a;  +  2. 

9.  4.x'-llx-3,  Sx^  +  6a^-llx'-23x-5. 

10.  2a;^-ar'^7/-4a?y4-3a;2/'j  S  x^y  -  10  x'y^  -{- 12  xf  - 10  y\ 

11.  6 m^— 17  mhi  —  1  mn^-\-^ n^,  12 m^  — 13  m^nH-21  myi^—Qn^. 

12.  2ar'4-5a;*-2ar^  +  3a^,  3  a;«  +  8a5^- 2a;*  +  a^- GojI 

13.  a^-2a^-2a?  +  la-Q,  d" -4.a^ ^d" ^1  a-2. 

It  follows  from  §  442  that,  if  two  expressions  are  prime  to 
each  other  (§  128),  their  product  is  their  L.  C.  M. 

443.  The  L.  C.  M.  of  three  expressions  may  be  found  as 
follows : 

Let  A,  B,  and  C  be  the  expressions. 

Let  M  be  the  L.  C.  M.  of  A  and  B ;  then  every  common  mul- 
tiple of  3f  and  (7  is  a  common  multiple  of  A,  B,  and  C. 

But  since  every  common  multiple  of  two  expressions  is  ex- 
actly divisible  by  their  L.  G.  M.,  every  common  multiple  of  A, 
B,  and  C  is  also  a  common  multiple  of  M  and  C. 

Then,  the  L.  C.  M.  of  M  and  C  is  the  L.  C.  M.  of  A,  B,  and  C. 

Hence,  to  find  the  L.  C.  M.  of  three  expressions,  find  the  L.  C.  M. 
of  two  of  them,  and  then  of  this  result  and  the  third  expression. 

We  proceed  in  a  similar  manner  to  find  the  L.  C.  M.  of  any 
number  of  expressions. 

EXERCISE  192 

Find  the  L.  C.  M.  of  the  following : 

1.   3a^-4a;-4,  3a;2-7x  +  2,  3  x^  - 10  a? -|- 8. 


404  ALGEBRA 

2.  2a«H-3a4-9a^  4a^  +  13a^  +  3a%  6  a^  +  13  a^-lS  a. 

3.  37i2-ll7i-4,  47i2-22w  +  24,  6?i2  +  lln  +  3. 

4.  4a3^_4a2-43a  +  20,  4  a3  +  20a2  +  13  a-12, 

4a«  +  12a2-31a-60. 

5.  2a^-5x  +  3,  4aj3-4a^  +  3a;-9,  4a^-13aj  +  6. 

444.  We  will  now  show  how  to  reduce  a  fraction  to  its  low- 
est terms,  when  the  numerator  and  denominator  cannot  be 
readily  factored  by  inspection. 

By  §  127,  the  H.  C.  F.  of  two  expressions  is  their  common 
factor  of  highest  degree,  having  the  numerical  coefficient  of 
greatest  absolute  value  in  its  term  of  highest  degree. 

We  then  have  the  following  rule : 

Divide  both  numerator  and  denominator  by  their  H.  C.  F. 

Ex.   Reduce  ^  a^  -  H  a2  + 7  a -6  ^^  .^^  lowest  terms. 
2  a-^  -  a  -  3 

By  the  rule  of  §  440,  we  find  the  H.  C.  F.  of  6  a^ -11  a"^ +  7  a -6  and 
2  a-2  -  a  -  3  to  be  2  a  -  3. 

Dividing  6  a^  _  n  ^^2  _,_  7  ^  _  6  \)y  2  a  -  3,  the  quotient  is  Sa^-a  +2. 
Dividing  2  a^  —  a  —  3  by  2  a  —  3,  the  quotient  is  a  +  1. 

6a3_iia2  +  7a_6     3a2-a  +  2 


Then, 


2a2-a-3  a  +  1 


EXERCISE  193 

Reduce  each  of  the  followins:  to  its  lowest  terms 


1. 

6a2_5a-4 

2. 

6  m2  -h  7  m  -  20 

4m2  +  16m  +  15 

3. 

5a^-lSab-6b' 

4a2-15a&  +  9&2 

4. 

4a^-4a;-3 

3a^-M0a;-f-8 


6. 


5a^  +  lla^-^22a-\~4: 


5a3  +  8a2  +  16a-8 

7     4yi«-6n^  +  Syt-6 
n^-Sn^-2n  +  4:  ' 


8. 


4a3  4.l3a26_4a&--6ft3 


4  0^3  _  4  ^2  _  5  ^  _  1  8  a^  4. 14  a5  -  15  ?/ 


MISCELLANEOUS  TOPICS  405 

9       ea^-o^-llx  +  Q  ^Q    2a^-9x'y-2xy'-15if 

'    9a^-lSx'-^nx-2'  '    2x^-7  ary-Uxy'+5f' 


PROOF  OF   (1),  §  235,   FOR  ALL  VALUES  OF  m  AND  n 


445.   I.    Let  m  =  ^  and  n  =  -,  where  p,  q,  r,  and  s  are  posi- 

re  integers.       p     ^r       ps       I-         __ 

We  have,        a^  x  a*  =  a«'  x  a''  =  Va^'  x  Vo^'"     (§  237) 


=  Va^"  X  a^'  (§  234)  =  Va^^+'''  (§  56)  =a  ''    (§  237)  =  a'  ». 

We  have  now  proved  that  (1),  §  235,  holds  when  m  and  w 
are  any  positive  integers  or  positive  fractions. 

II.  Let  m  be  a  positive  integer  or  fraction ;  and  let  n  =  —  q, 
where  g  is  a  positive  integer  or  fraction  less  than  m. 

By  §§  56,  or  445,  I,  a*"-*  x  a'  =  a"^-^-^^  =  a"*. 

Whence,  a'"-«=  —  =  a'"  x  a"'  (§  240). 

a' 

That  is,  a'"  x  a~'  =  a'""''. 

III.  Let  m  be  a  positive  integer  or  fraction ;  and  let  n  =  q, 
where  g'  is  a  positive  integer  or  fraction  greater  than  m. 

By  §  240,  a'"  x  a-^=  ^—  =  -^—  (§  445,  II)  =  a*""'. 

IV.  Let  m  =  —p  and  n  =  —  q,  where  p  and  g  are  positive  inte- 
gers or  fractions. 

Then,  a"^  x  a"'?  ==  —  =  —  (§  §  56,  or  445,  I)  =  a"^"'. 

Then,  a"*  x  a"  =  a"*"^"*  for  all  positive  or  negative,  integral  or 
fractional,  values  of  m  and  n.         . 

446.  We  will  now  show  how  to  reduce  a  fraction  whose 
denominator  is  irrational  to  an  equivalent  fraction  having  a 
rational  denominator,  when  the  denominator  is  the  sum  of  a 
rational  expression  and  a  surd  of  the  nth  degree,  or  of  two 
surds  of  the  ?ith  degree. 


406  ALGEBRA 


1.   Reduce    -^   to  an  equivalent    fraction    having    a 

2  +  ^3  ,  ^ 

rational  denominator. 

1  1 

We  have, 


2  +  v^     8^  +  3^ 
Now,  (a  +  &)  (a2  -  ab  +  b^)  =  a^ -\- b^  (§  102). 

Then,  if  we  multiply  both  terms  by  8^  -  83  .  S^  +  3^,  the  denomiDator 
will  become  rational ;  thus, 

1       _  8^  -  8^  ■  3^  +  3^  _  (8^)2  -  8^  .  33  +  3^ 

8^  +  3^     (8^ +  3^)  (8^ -8^.  33 +  3^)  8  +  3 

4-2^/3  +  v'9 


2.    Reduce —   to  an  equivalent  fraction   haviner  a 

</T-</E 

rational  denominator. 

1  1 

We  have, 


^_^5      7^-5^ 
Now,        (a  -  6)  (a3  +  a%  +  ab^  +  b^)  =  a*  -  6*  (§  103). 
Then,  if  we  multiply  both  terms  by 

the  denominator  will  become  rational ;  thus, 

1  7^  +  7^  .  5*  +  7^  .  5^  +  6* 


7i  _  5?      (7?  _  6?)  (7?  +  7^  .  6^  +  7*  •  5^  +  6^) 

_  \/7«  +  V^72  .  5  +  VTT52  +  \/p  ^  v/343  +  v^245  +  Vm  +  -C/125 
7-5  ~  2 

The  .method  of   §  446  can  be  applied  to  cases  where  the 
denominator  is  in  the  form  Va-|-V&,  or  Va  —  -\/b. 

3.   Reduce to  an  equivalent  fraction  having  a 

rational  denominator. 

The  lowest  common  multiple  of  the  indices  3  and  2  is  6. 


MISCELLANEOUS  TOPICS  407 

We  have,         -—z =  -— z — rzz  = ; 7' 

y/2  +  VE      ^22  +  v^53      (22)i  +  (53)i 

Now,       (a  +  b)  (a5  _  a*^  +  QjS^a  _  ^253  ^  0^54  _  55)  =  a^_  ^6. 

Then,  if  we  multiply  both  terms  by 

(22) f  _  (22)t(53)^  +  (2-2)1(53)1  -  (22)i(53)t  +  (22)i(53)t  -(53)t, 

the  denominator  will  become  rational ;  thus, 

1  _  2t  -  2t  ■  5t  4-  2  •  5  -  2t  •  5t  +  2i  •  5^  -  5t 

(22)i+(53)5  (22)t  +  (53)t 

_  2  ♦  2!  -  2  >  2t .  5t  +  10  -  2^  .  5  «  5t  +  2^  ♦  52  -  52  .  5I 
22  +  53 

_2v^-2v^^2T58  +  l0-5v^^*^  +  25\/2-25\/5 
~  4  +  125 

_10  +  2v^i-2v^500-5\/2000  +  25v^2-25\/5 
129 

EXERCISE  194 

Keduce  each  of  the  following  to  an  equivalent  fraction  having 
a  rational  denominator : 

1 


1.  J-  3    ___L__.  5 

2    ^_.  4.   ^ 6.   V3-V2. 

2--V/4  a/3  4- a/4  V3  +  V2 


THE  BINOMIAL  THEOREM  FOR  POSITIVE  INTEGRAL 
EXPONENTS 

447.  In  the  proof  of  §  387,  we  only  considered  the  first  four 
terms  of  the  expansion  of  (a  +  a;)*'+^,  in  equation  (2). 

To  make  the  proof  complete,  we  must  show  that  the  fifth  law 
of  §  386  holds  for  any  two  consecutive  terms,  in  equation  (2). 

Let  P,  Q,  and  R  denote  the  coefficients  of  the  terms  involv- 
ing a'»-''a^",  a'*~''~V+\  and  a""''~V+2,  respectively,  in  the  second 
member  of  (1),  §  386. 


408  ALGEBRA 

Thus,  (a  +  «)"  =  a"  +  na^'-'^x  +  •  •  • 

+  Pa^-'af  +  Qa''-'-- V+i  +  i2a"-'-V+2  +  ....  (3) 

Multiplying  both  members  by  a  +  x,  we  have 

(a  +  xy+^  =  a'^+i  +  na^'x  -\ \-  Qa"-'"x'-+i  +  Ra^-'-^xr^^  +  ... 

+   a'^ic  H h  Pa'-'-iC+i  +  Qa"-*- V+2  +  . . . 

=  a'*+i  +  (7i4-l)a"a;+-.. 

+  (P4-  Q)a'*-'-aj'-+i+(Q+i2)a"-'-V+2+ ....      (4) 

Since  the  fifth  law  of  §  386  is  assumed  to  hold  with  respect 
to  the  second  member  of  (3),  we  have 

r  +  1  r  +  2 
Therefore, 

Q{n-T-1)  Qin  +  V) 

Q  +  E                    r  +  2  r^'l         n-r 


P+Q  Q(r-hl)  I  Q         Q(n  +  1)      r  +  2 

n—r  n—r 

Whence,  Q^]i=(p^Qy 


n  —  r 

n  —  r 


r  +  2 

But  n  —  r  is  the  exponent  of  a  in  that  term  of  (4)  whose  coeffi- 
cient is  P-^Q,  and  r  +  2  is  the  exponent  of  x  increased  by  1. 

Therefore,  the  fifth  law  holds  with  respect  to  any  twp  con- 
secutive terms  in  equation  (2),  §  387. 


wp 

/I 


THE  THEOREM  OF  UNDETERMINED  COEFFICIENTS 

448.  Before  giving  the  more  rigorous  proof  of  the  Theorem 
of  Undetermined  Coefficients,  we  will  prove  two  theorems  in 
regard  to  infinite  series. 

First,  if  the  infinite  series 

a -\-hx  +  cx^ -{- da^ -\- "* 

is  convergent  for  some  finite  value  of  x,  it  is  Jinite  for  this  value 
of  X  (§  393),  and  therefore  finite  when  x=0. 
Hence,  the  series  is  convergent  when  a;  =  0. 


MISCELLANEOUS  TOPICS  409 

449.  Second,  if  the  infinite  series 

is  convergent  for  some  finite  value  of  x,  it  equals  0  when  a;  =  0. 

For,  ax  +  6a^  +  ca;^  -|-  •  •  •  is  finite  for  this  value  of  x,  and 
hence  a  +  6aj  +  ca^  +  •••  is  finite  for  this  value  of  x. 

Then,  a  -\- hx -{-  cx^  -\-  •  •  •  is  finite  when  aj  =  0 ;  and  therefore 
x{a-\-hx  +  C7?  +  •••),  or  ax  +  hy?-\-CQi^'-\-  •••,  equals  0  when  a;=0. 

450.  Proof  of  the  Theorem  of  Undetermined  Coefficients  (§  396). 
The  equation 

^  +  ija;  +  (7x2  +  Z>a^  +  ...  =  ^'  4.  B'x  +  Ox^  +  Z)'a^  +  ...  (1) 

is  satisfied  when  x  has  any  value  which  makes  both  members 
convergent ;  and  since  both  members  are  convergent  when  x  =  0 
(§  448),  the  equation  is  satisfied  when  x  =  0. 

Putting  X  =  0,  we  have  by  §  449, 

5x  +  (7x2  +  i)x3  +  ...  ^  0,  and  B'x  +  C'x^  +  D'x^  +  ...  =  0. 

Whence,  A  =  A'. 

Subtracting  A  from  the  first  member  of  (1),  and  its  equal  A' 
from  the  second  member,  we  have 

Bx -\- Cx"  +  Da^  +  ...  =  B'x  +  C'x^  +  D'^  +  •-. 

Dividing  each  term  by  x, 

B  -{.  Cx  -h  Dx'  +  •"  =  B'  -{-  C'x  +  D'x'  +  ....  (2) 

The  members  of  this  equation  are  finite  for  the  same  values 
of  X  as  the  given  series  (§  449). 

Then,  they  are  convergent,  and  therefore  equal,  for  the  same 
values  of  x  as  the  given  series. 

Then  the  equation  (2)  is  satisfied  when  x  =  0. 

Putting  x  =  0,  we  have  B=B'. 

Proceeding  in  this  way,  we  may  prove  C  =  C,  etc. 


410  ALGEBRA 


XXXIII.     THE    FUNDAMENTAL    LAWS    FOR 
ADDITION  AND  MULTIPLICATION 

451.  The  Commutative  Law  for  Addition. 

If  a  man  gains  ^  8,  then  loses  $  3,  then  gains  $  6,  and  finally 
loses  $  2,  the  effect  on  his  property  will  be  the  same  in  what- 
ever order  the  transactions  occur. 

Then,  with  the  notation  of  §  16,  the  result  of  adding  +  $  8, 
—  $  3,  +  $  6,  and  —  $2,  will  be  the  same  in  whatever  order 
the  transactions  occur. 

Then,  omitting  reference  to  the  unit,  the  result  of  adding 
+  8,  —  3,  +6,  and  —  2  will  be  the  same  in  whatever  order  the 
numbers  are  taken. 

This  is  the  Commutative  Law  for  Addition,  which  is : 

The  sum  of  any  set  of  numbers  will  he  the  same  in  whatever 
order  they  may  he  added. 

452.  The  Associative  Law  for  Addition. 

The  result  of  adding  6  +  c  to  a  is  expressed  a  +  (6  -f  c),  which 
equals  (&+c)-|-a  by  the  Commutative  Law  for  Addition  (§  451). 

But  (&  +  c)  -f-  a  equals  6  +  c  +  a,  by  the  definition  of  §  3 ;  and 
h-\-c-{-a  equals  a+h-\-c,  by  the  Commutative  Law  for  Addition. 

Whence,  .  a -\-  (h  +  c)  =  a -\-h -\- c. 

Then,  to  add  the  sum  of  a  set  of  numbers,  we  add  the  num- 
bers separately. 

This  is  the  Associative  Law  for  Addition. 

453.  The  Commutative  Law  for  Multiplication. 

The  product  of  a  set  of  numbers  will  be  the  same  in  whatever 
order  they  may  be  multiplied. 

By  §  55,  the  sigji  of  the  product  of  any  number  of  terms  is 
independent  of  their  order ;  hence,  it  is  sufficient  to  prove  the 
commutative  law  for  arithmetical  numbers. 


LAWS  FOR  ADDITION  AND  MULTIPLICATION    411 

Let  there  be,  in  the  figure,  a  stars  in  each  row,  a  m  2^  row 
and  h  rows.  *  #  #  #  .. 

We  may  find  the  entire  number  of  stars  by  *  *  *  ^  -• 
multiplying  the  number  in  each  row,  a,  by  the  *  *  *  ^  •• 
number  of  rows,  h.  '  '  ' 

Thus,  the  entire  number  of  stars  is  a  x  6. 

We  may  also  find  the  entire  number  of  stars  by  multiply- 
ing the  number  in  each  vertical  column,  b,  by  the  number  of 
columns,  a. 

Thus,  the  entire  number  of  stars  is  b  x  a. 

Therefore,  a  xb  =  b  x  a. 

This  proves  the  law  for  the  product  of  two  positive  integers. 
Again,  let  c,  d,  e,  and/  be  any  positive  integers. 

C        S  C  X  s 

Then,   -  x  —  = ;    for,   to   multiply  two   fractions,   we 

multiply  the  numerators  together  for  the   numerator  of  the 
product,  and  the  denominators  together  for  its  denominator. 

Then,  -  x  —  = ;  since  the  commutative  law  for  multi- 

'  d      f      fxd' 

plication  holds  for  the  product  of  two  positive  integers. 

C  S  6  C 

Hence,  -  x  —  =  —  x  -;  which  proves  the  commutative  law 
(^     f      f     d 

for  the  product  of  two  positive  fractions. 

454.   Thei  Associative  Law  for  Multiplication. 

To  midtiply  by  the  product  of  a  set  of  riumbers,  we  multiply  by 
the  numbers  of  the  set  separately. 

This  law  was  assumed  to  hold  in  §§  56  and  57. 

The  result  of  multiplying  a  by  be  is  expressed  a  x  (be),  which 
equals  (be)  X  a,  by  the  Commutative  Law  for  Multiplication. 

But  by  the  definition  of  §  5,  (be)  x  a  equals  bca,  which  equals 
abc  by  the  Commutative  Law  for  Multiplication. 

Whence,  a  x  (be)  =  abc. 

This  proves  the  law  for  the  product  of  three  numbers. 


412  ALGEBRA 

The  Commutative  and  Associative  Laws  for  Multiplication  may  be 
proved  for  the  product  of  any  number  of  arithmetical  numbers. 
(See  the  author's  Advanced  Course  in  Algebra,  §§  18  and  19.) 

455.   The  Distributive  Law  for  Multiplication. 

The  law  is  expressed  (a  +  h)c  =  ac-\-bc  (§  40). 

We  will  now  prove  this  result  for  all  values  of  a,  b,  and  c. 

I.  Let  a  and  b  have  any  values,  and  let  c  be  a  positive 
integer. 

Then,  (a  +  b)c  =  {a  +  b) -{- (a -{- b) -\-  •••  to  c  terms 

=  (a-{-a-\-  '••  to  c  terms)  +  (6  -f-  &  +  •  •  •  to  c  terms) 
(by  the  Commutative  and  Associative  Laws  for  Addition), 
=  ac  -f-  be. 

II.  Let  a  and  b  have  any  values,  and  let  c  =  — ,  where  e  and 
/  are  positive  integers.  ^ 

Since  the  product  of  the  quotient  and  divisor  equals  the 
dividend,  ^ 

Then,        (a  +  b)  x  jX  f=(a-\-b)  xe  =  ae-^be,  by  I. 

Whence,  (a  +  b)  x  ^X  f=a  x -,  x  f+b  x  ^  X  f. 
J  J  J 

Dividing  each  term  by  /  (Ax.  8,  §  9),  we  have 

(a  +  6)x^  =  ax-^  +  6x-. 

Thus,  the  result  is  proved  when  c  is  a  positive  integer  or  a 
positive  fraction. 

III.  Let  a  and  b  have  any  values,  and  let  c  =  —  g,  where  g  is 
a  positive  integer  or  fraction. 

By  §  54,  (a  +  5)(-  (7)  =  -  (a  +  b)g^  -  (ag  +  bg),  by  I  and  II, 

=  -ag-bg  =  a(-g)-^b(-g).       . 

Thus,  the  distributive  law  is  proved  for  all  positive  or  nega- 
tive, integral  or  fractional,  values  of  a,  b,  and  c. 


ADDITIONAL  METHODS  IN  FACTORING  413 


XXXIV.    ADDITIONAL    METHODS   IN 
FACTORING 


456.   The  Remainder  Theorem. 

Let  it  be  required  to  divide  px^  -\-qx-\-  rhj  x 

py?  -\-    qx-\-r 
psi?  —  apx 


px  +  {ap  +  q) 
(ap  +  q)x  J 

(ap  +  q)x  —  pa^  —  qa 

pa^  -\-  qa-{-r,  Remainder. 

We  observe  that  the  final  remainder, 

pa^  -j-  qa-\-  r, 

is  the  same  as  the  dividend  with  a  substituted  in  place  of  x ; 
this  exemplifies  the  following  law : 

If  any  polynomial,  involving  x,  be  divided  by  x  —  a,  the 
remainder  of  the  division  equals  the  result  obtained  by  substi- 
tuting a  for  X  in  the  given  polynomial. 

This  is  called  The  Remainder  Theorem. 

To  prove  the  theorem,  let 

px""  +  5af~^  H \-rx-{-s 

be  any  polynomial  involving  x. 

Let  the  division  of  the  polynomial  hj  x  —  a  be  carried  on 
until  a  remainder  is  obtained  which  does  not  contain  x. 

Let  Q  denote  the  quotient,  and  R  the  remainder. 

Since  the  dividend  equals  the  product  of  the  quotient  and 
divisor,  plus  the  remainder,  we  have 

Q(x  —  a)  -\-  R  —  px""  4-  qx""'^  H \-rx-\-s. 

Putting  X  equal  to  a,  in  the  above  equation,  we  have, 

R  =  pa""  +  qa""-^  -\ \-ra-\-  s. 


414  ALGEBRA 

457.  The  Factor  Theorem. 

If  any  polynomial,  involving  x,  becomes  zero  when  x  is  put 
eqital  to  a,  the  polynomial  has  x  —  a  as  a  factor. 

For,  by  §456,  if  the  polynomial  is  divided  by  x  —  a,  the 
remainder  is  zero. 

458.  Examples.  f 

1.  Find  whether  cc  —  2  is  a  factor  of  a^  —  5  a^  +  8. 

Substituting  2  for  x,  the  expression  x^  —  bx^  -\-  S  becomes 
23- 5- 22 +  8,  or  -4. 

Then,  by  §  456,  if  x^  —  5x^  +  8  be  divided  by  x  —  2,  the  remainder  is 
—  4  ;  and  x  —  2  is  not  a  factor. 

2.  Find  whether  m  +  n  is  a  factor  of 


m 


4  m^n  +  2  m^n^  +  5  mn^  —  2  ?i^.  (1) 

Putting  m  =  —  w,  the  expression  becomes 

w*  +  4  w*  +  2  w*  -  5  w*  -  2  w*,  or  0. 

Then,  by  §  456,  if  the  expression  (1)  be  divided  by  m  +  n,  the  re- 
mainder is  0  ;  and  w  +  w  is  a  factor. 

3.  Prove  that  a  is  a  factor  of 

(a-\-b-\-c)  (ab  +  bc-\-  ca)  -  (a  +  6)  (6  +  c)  (c  -f-  a). 
Putting  a  =  0,  the  expression  becomes 

(6  +  6)1)0  -h{h-\-  c)c,  or  0. 
Then,  by  §  456,  a  —  0,  or  a,  is  a  factor  of  the  expression. 

4.  Factor  a^-3a;2-14a;-8. 

The  positive  and  negative  integral  factors  of  8  are  1,  2,  4,  8,  —  1,  —  2, 
—  4,  and  —  8. 

It  is  best  to  try  the  numbers  in  their  order  of  absolute  magnitude. 

If  X  =  1,  the  expression  becomes  1—3—14  —  8. 

If  X  =  —  1,  the  expression  becomes  —1  —  3  +  14  —  8. 

If  X  =  2,  the  expression  becomes  8  —  12  —  28  —  8. 

If  X  =  —  2,  the  expression  becomes  —  8  —  12  +  28  —  8,  or  0. 

This  shows  that  x  +  2  is  a  factor. 

Dividing  the  expression  by  x  +  2,  the  quotient  is  x^  —  5  x  —  4. 

Then,  x^  -  3x2  -  14  x  -  8  =(x  +  2)(x2  -  5x  -  4). 


<^  1 

ADDITIONAL  METHODS  IN   FACTORING  416 

EXERCISE  195 

Factor  the  following : 
1.    a^  +  1.  2.   0^^-81.  3.   a;«-64. 

4.  a;^  +  4a^  +  7a;-12.  8.    a^-18aj  +  8. 

5.  x''-Q(?-\-Qfx'  +  l^x-\-Q>.        9.    a;3-5cc2_8ic  +  48. 

6.  a?3-x2-llx-10.  10.    a^4_^8ar^  +  13a^'-13a;-4. 

7.  ar^-9a^  +  15£c  +  9.  11.    x^ ^Qix" -x-30. 

Find,  without  actual  division, 
.-^  12.    Whether  a;  -  3  is  a  factor  of  ar"  -  6  x"  + 13  a;  - 12.      :/ >-i 

13.  Whether  a?  4-  2  is  a  factor  of  ar'  +  7  a^  —  6. 

14.  Whether  a;  + 1  is  a  factor  of  a;''  —  4a^  +  2a52  —  2a5  —  9. 
^15.    Whether  a;  is  a  factor  of  x{y  +  zf  +y{z  +  xf  +  z{x  +  yf. 

16.  Whether  a  is  a  factor  of  a\b-cf-[-h\c-af+c\a-hf. 

17.  Whether  a;  —  ?/  is  a  factor  of  (a;  —  yf  +  (2/,—  =2)^  +  (2  —  xf. 

18.  Whether  m  H-  n  is  a  factor  of  m{m  -^  2  Tif  —  n{2  m -\-  nf. 

459.   We  will  now  give  formal  proofs  of  the  statements  of 
§  104. 

Proof  of  1. 

If  h  be  substituted  for  a  in  a''  —  6",  the  result  is  6"  —  6",  or  0. 

Then,  by  §  457,  a"  —  6"  has  a  —  Z>  as  a  factor. 

Proof  of  II. 

If  —h  be  substituted  for  a  in  a^'—b'',  the  result  is  (— ^)"— 6"; 
or,  since  w  is  even,  b^'—b'*,  or  0. 

Then,  by  §  457,  a"  —  6"  has  a  +  6  as  a  factor. 

Proo/  of  III. 

If  -6  be  substituted  for  a  in  a'^+ft^the  result  is  (-6)" +6"; 
or,  since  n  is  odd,  —  &"  -I-  6",  or  0. 
Then,  a"  4-  6"  has  a  +  6  as  a  factor. 


416  ALGEBRA 

Proof  of  ly. 

If  —  6  or  -f  6  be  substituted  for  a  in  a"  +  6",  the  results  are 
(—&)"  +  ^"  01*  &"  +  ^"^  respectively. 

Since  n  is  even,  neither  of  these  is  zero. 

Then,  neither  a  +  6  nor  a  —  6  is  a  factor  of  a"  +  6". 

SYMMETRY 

460.  An  expression  containing  two  or  more  letters  is  said  to 
be  symmetrical  with  respect  to  them,  when  any  two  of  thepi 
can  be  interchanged  without  altering  the  value  of  the  expres- 
sion. 

Thus,  ah-{-hc-\-  ca  is  symmetrical  with  respect  to  the  letters 
a,  6,  and  c ;  for  if  a  and  b  be  interchanged,  the  expression  be- 
comes ba  +  ac  i-  cb,  which  is  equal  to  ab  -\-bc-\-  ca. 

And,  in  like  manner,  the  expression  is  not  altered  in  value  if 
we  interchange  b  and  c,  or  c  and  a. 

461.  Cyclo-symmetry. 

An  expression  containing  n  letters  a,  b,  c,  ">,  m,  n,  is  said  to 
be  cyclo-symmetrical  with  respect  to  them  when,  if  a  be  replaced 
by  b,  bhj  c,  •••,  m  by  n,  and  n  by  a,  the  value  of  the  expression 
is  not  changed. 

The  above  is  called  a  cyclical  interchange  of  letters. 

Thus,  the  expression  a^b  +  b^c  +  c^a  is  cyclo-symmetrical  with 
respect  to  the  letters  a,  b,  and  c ;  for  if  a  be  replaced  by  b,  b  by  c, 
and  c  by  a,  the  expression  becomes  b^c  +  c'^a  +  a^6,  which  is  equal 
to  a^b  +  b^c  -{-  c^a. 

The  above  expression  is  not  symmetrical  with  respect  to  a,  6,  and  c ; 
for  if  a  and  h  be  interchanged,  the  expression  becomes  b^a  -\-  a^c  +  d^b, 
which  is  not  equal  to  a^h  +  b'^c  +  c^a. 

462.  It  follows  from  §§  460  and  461  that,  if  two  expressions 
are  symmetrical  or  cyclo-symmetrical,  the  results  obtained  by 
adding,  subtracting,  multiplying,  or  dividing  them  are,  respec- 
tively, symmetrical  or  cyclo-symmetrical. 


ADDITIONAL  METHODS  IN  FACTORING  417 

463.  Applications. 

The  principle  of  symmetry  is  often  useful  in  abridging  alge- 
braic operations. 

1.  Expand  (a  +  b-^cy. 

We  have,  (a  -\-  b  +  cy  =  {a  -\-  b  +  c)(a  +  b  +  c)(a  +  b  -{-  c). 

This  expression  is  symmetrical  with  respect  to  a,  6,  and  c  (§  460) ,  and 
of  the  third  degree. 

There  are  three  possible  types  of  terms  of  the  third  degree  in  a,  6, 
c ;  terms  like  a^,  terms  like  a%,  and  terms  like  abc. 

It  is  evident  that  a^  has  the  coefficient  1  ;  and  so,  by  symmetry,  b^  and 
c^  have  the  coefficient  1. 

The  a^b  terms  may  be  obtained  by  multiplying  the  a's  in  any  two  fac- 
tors by  the  b  in  the  remaining  factor. 

Then,  it  is  evident  that  a-b  has  the  coefficient  3  ;  and  so,  by  symmetry, 
have  b'^a,  b^c,  c^b,  c^a,  and  a^c. 

Let  m  denote  the  coefficient  of  abc. 

Then,  («  +  6  +  c)3 

=  a^ -i- b^  +  c^ -{-  S(a~b  -\-  b'^a  +  b'^c  +  c%  +  d^a  +  a'^c)  +  mabc. 

To  determine  m,  we  observe  that  the  above  equation  holds  for  all  values 
of  a,  b,  and  c. 

We  may  therefore  let  a  =  ?>  =  c  =  1. 

Then,  27  ==  3  +  18  +  m  ;  and  m  =  Q. 

Whence,  (a  +  b  +  cy 

=  a^  +  &3  4.  c3  +  3(a26  +  &%  +  b^^  +  c2&  +  c^a  +  «%)  +  6  abc. 

2.  Expand  (x  —  y  —  zy  -\-  (y  —  z  —  xy  -\-  (z  —  x  —  yy. 

This  expression  is  symmetrical  with  respect  to  x,  y,  and  z,  and  of  the 
second  degree. 

The  possible  types  of  terms  of  the  second  degree  in  x,  y,  and  z  are 
terms  like  x^,  and  terms  like  xy. 

It  is  evident,  by  the  rule  of  §  204,  that  x^  has  the  coefficient  3 ;  and  so, 
by  symmetry,  have  y^  and  z^. 

Let  m  denote  the  coefficient  of  xy. 

Then,  (x- y  -  z)'^ -\- (y  -  z  -  xy +{z  -  x  -  yy 

=  S(x'^ +  y^ -{-z^)+ m(xy-^yz-{-zx). 

To  determine  m,  put  x  =  y  =  z  =  l. 
Then,  3  =  9 +  Sm,  or  m  =  -2. 

Whence,  (x  —  y  —  zy  -\-(y  —  z  —  xy  -^(z  —  x  —  yy 

=  3(x2  +  2/2  +  s2)  -  2(xy  +  yz  +  zx). 


418  ALGEBRA 

3.   Expand 

(a  +  6  +  c)3  +  (a  +  &  -  c)3  +  (5  +  c  -  a)3  +  (c  +  a  -  by. 

The  expression  is  symmetrical  with  respect  to  a,  6,  and  c,  and  of  the 
third  degree. 

The  possible  types  of  terms  are  terms  like  a^,  terms  like  a^b,  and  terms 
like  abc. 

It  is  evident,  by  proceeding  as  in  Ex.  1,  that  a^  has  the  coeflacient 
1_|.1_14-1,  or2;  and  so,  by  symmetry,  have  b^  and  c^. 

Again,  proceeding  as  in  Ex.  1,  it  is  evident  that  a^b  has  the  coefficient  3 
in  the  first  term,  3  in  the  second,  3  in  the  third,  and  —  3  in  the  fourth. 

Then,  a'^b  has  the  coefficient  3+3  +  3-3,  or  6;  and  so  by  symmetry 
have  b-a,  b%  c^b,  d^a,  and  a^c. 

Let  m  denote  the  coefficient  of  abc.        — 

Then,  (a  +  6  +  c)3  +  (a  +  6  -  c)3  +  (&  +  c  -  a)^  +  (c  +  a  -  6)8 

=  2{a^  +  b'^-\-  c3)  +  6(a26  +  &%  +  b-c  +  c^b  +  c-'a  +  a^c)  +  mabc. 

To  determine  m,  let  a  =  &  =  c  =  1. 

Then,  27  +  1  +  1  +  1  =  6  +  36  +  m,  or  m  =  -  12. 

Then,   (a  +  6  +  c)^  +  (a  +  6  -  c)3  +  (6  +  c  -  ay  +  (c  +  a  -  by 

=  2(a3  +  63  +  c3)  +  6(^25  +  b^a  +  b'^c  +  0^6  +  c^a  +  a'^c)  -  12  abc. 

EXERCISE  196 

1.  In  the  expansion  of  an  expression  whicli  is  symmetrical 
with  respect  to  a,  b,  and  c,  what  are  the  possible  types  of  terras 
of  the  fourth  degree  ?  of  the  fifth  degree  ? 

2.  If  one  term  of  an  expression  which  is  symmetncal  with 
respect  to  a,  b,  and  c  is  (2  a  —  b  —  c)(2  b  —  c  —  a),  what  are 
the  others? 

3.  Is  the  expression  a(6  —  c)^4- 6(c  — a)^  +  c(a  — &)^  sym- 
metrical with  respect  to  a,  b,  and  c? 

4.  Is  the  expression  (sc^  —  y'^y  +  (y^  —  z^y  +  (z^  —  x^^  sym- 
metrical with  respect  to  x,  y,  and  z  ? 

Expand  the  following  by  the  symmetrical  method  : 
5.    (a  +  b  +  cy.  6.    (a  +  6  +  c  +  d)l 


ADDITIONAL  METHODS  IN  FACTORING  419 

7.  (x  +  y-zy^{y  +  z-xy^{z  +  x-y)\ 

8.  (2  a  -  3  6  -  4  c)2  +  (2  &  -  3  c  -  4  a)2  +  (2  c  -  3  a  -  4  hf. 

9.  (a  4-  6  4-  c)3  -f  (a  -  6  -  c)'^  +  (6  -  c  -  a)^  +  (c  -  a  -  6)^. 

10.  {a-\-h^c-df-\-{h  +  c-{-d-ay  +  (G  +  d-^a-hY 

-\-{d-\-a  +  h  —  cf. 

11.  (aH-6  +  c  +  (?)^ 

12.  (x  +  2/ -2;)(i/H-2;-a;)(2  +  i»-?/). 

13.  (a  +  &  +  c)(a  +  h-—  c)(b  -\-c-  a)(c  -\-a  —  b). 

14.  (a^-^y^  +  z^-\-2xy  +  2yz  +  2zxy. 

464.   Factoring  of  Symmetrical  Expressions. 

The  method  of  §  457  is  advantageous  in  factoring  symmet- 
rical expressions  (§§  460,  461). 

1.  Factor 

a(b  +  cf  +  5(c  +  a)2  +  c(a  +  bf  -  a\b  +  c)  -  bXc  +  a)  -  c^a  +  &)• 

The  expression  is  symmetrical  with  respect  to  a,  6,  and  c. 

Being  of  the  third  degree,  the  only  literal  factors  which  it  can  have  are 
three  of  the  type  a ;  three  of  the  type  «  +  6;  ora  +  6  +  c,  and  a  factor 
of  the  second  degree. 

Patting  a  =  0,  the  expression  becomes 

6c2  +  c62  -  h^c  -  c%,  or  0. 

Then,  by  §  457,  a  is  a  factor  ;  and,  by  symmetry,  h  and  c  are  factors. 
The  expression,  being  of  the  third  degree,  can  have  no  other  literal  fac- 
tor; but  it  may  have  a  numerical  factor. 
Let  the  given  expression  =  mahc. 
To  determine  m,  let  a  =  6  =  c  =  1. 

Then,  4  +  4  +  4-2-2-2  =  wi,  or  w  =  6. 

Whence,  the  given  expression  =  6  dhc. 

2.  Factor  x^ -\- if -\- z^  —  ^  xyz. 

The  expression  is  symmetrical  with  respect  to  cc,  ?/,  and  z. 
The  only  literal  factors  which  it  can  have  are  three  of  the  type  x ; 
three  of  the  type  x  +  y\  or  x -\- y  -\-  z,  and  a  factor  of  the  second  degree. 


; 


420  ALGEBRA 

It  is  evident  that  neither  x^  y,  nor  0  is  a  factor. 
Putting  X  equal  to  —  y,  the  expression  becomes 

-y^  +  y^  +  z^  +  'S  x% 
which  is  not  0. 

Then,  x -{•  y  is  not  a  factor  (§  457)  ;  and,  by  symmetry,  neither  y  +  z 
nor  z  +  X  is  a  factor. 

Putting  X  equal  to  —  y  —  z,  the  expression  becomes 

{-y  -  zY  +  y^  +  z^  -^{-y  -  z)yz 

=  -y^  -  Sy^z-Syz^  -  z^  +  y^-h  z^  +  3  y'^z  +  3  yz^  =  0. 

Therefore,  x  +  y  -\-  z  is  a  factor. 

The  other  factor  may  be  obtained  by  division,  or  by  the  following 
process  : 

It  is  of  the  second  degree  ;  and  as  it  is  symmetrical  with  respect  to  x, 
y,  and  z,  it  must  be  of  the  form 

m(x2  +  2/2  +  z^)  +  n{xy  +  yz  +  zx). 

It  is  evident  that  w  =  1,  as  this  is  the  only  value  which  will  give  the 
terms  x^,  y^,  and  z^  in  the  given  expression. 
Then, 

a;3  +  2/3  4.  2-3  _  3 xyz=(x-\-y  +  z) [x2  +  y2  ^  z"^  ^  n  (xy  +  yz  +  zx)^. 
To  determine  7i,  let         x  =  I,  y  =  I,  z  =  0. 

Then,  2  =  2(2  +  w),  or  1=2  +  w,  or  n  =  -  1. 

Whence, 
x^  -\- y^  -\-  z^  —  Z  xyz  =  (x-\-  y  +  z)  (x^  -\-  y"^  -\-  z"^  -  xy  -  yz  -  zx). 

3.   Factor  ab(a  —  b)  -\- be  {h  —  c) -\-  ca(G  —  a). 

The  expression  is  cyclo-symmetrical  (§  461)  with  respect  to  a,  &,  and  c. 

It  is  evident  that  neither  a,  6,  nor  c  is  a  factor. 

The  expression  becomes  0  when  a  is  replaced  by  b. 

Then,  a  —  6  is  a  factor ;  and,  by  symmetry,  b  —  c  and  c  —  a  are 
factors. 

The  expression  can  have  no  other  literal  factor,  but  may  have  a  numeri- 
cal one. 

Let  the  given  expression       =  m(a  —  b)(b  —  c)(c  —  a). 

To  determine  m,  let  a  =  2,  &  =  1,  and  c  =  0. 

Then,  2  =  —  2  m,  and  m  =  —  1. 

Then,  the  given  expression  =  —  (a  —  b)(b  —  c) (c  —  a). 


ADDITIONAL  METHODS  IN  FACTORING  421 

EXERCISE  197 
Factor  the  following : 

1.  m»  +  2m2n4-2mri2  +  n3. 

2.  (ab-\-bc-^ca)(a+b+c)-a\b+c)-b%c-\-a)-c\a  +  b). 
3    x'(y-^z)+y\z  +  x)-\-z\x-^y)-{-2xyz. 

4.  a(b  +  cy  +  b(c  4-  a.)2  +  c(a  +  6)^-4  a6c. 

5.  a\b-c)-^b\c-a)-\-c\a-b). 

6.  (a;  +  2/  +  2)(a;.v  +  2/2:  +  z^)  -(x  +  y)(y  +  ^)(^  +  a?)- 

7.  a6(a  +  6)  +  6c(6  +  c)  -f-  ca{c  +  a)  +  2  a6c. 

8.  (a;  +  2/  +  2)3_a^_2/3_^ 

9.  (x  +  y-{-z)(xy-{-yz-{-zx)-xyz. 

10.  (a;-y)3  4-(2/-;2)3  +  (2-aj)3. 

11.  a\b  -  c)  +  &'(c  -  a)  +  c3(a  -  6). 


422  ALGEBRA 


XXXV.    MATHEMATICAL  INDUCTION 

465.  In  §  387  we  gave  an  example  of  Mathematical  Induc- 
tion, in  proving  the  Binomial  Theorem  for  a  Positive  Integral 
Exponent ;  in  the  present  chapter,  we  will  give  other  illustra- 
tions of  the  method. 

466.  AVe  will  now  prove  that  the  laws  of  §  103  hold  univer- 
sally. 

We  will  first  prove,  by  Mathematical  Induction,  that  they 

hold  for ,  where  n  is  any  positive  integer. 

Assurme  the  laws  to  hold  for  ,  where  n  is  any  positive 

.   ,  a  —  b 

integer. 

Then,  ^^-^  =  a»-i  -f  a^-%  H-  c^-^h^  +  •  •  -  +  h^-\  (1) 

a  —  h 

a  —  h  a—  h 

^  a\a  —  h)  -h  hjal'  —  6^) 
a~h 

=  a""  +  6(a"-^  -L  a'^-26  -i-  a''-^^  -\ \-  6"-^),  by  (1), 

=  a"  _l_  a"-i5  ^  cjt^  -2^2  ^  ...  j^  ^n 

This  result  is  in  accordance  with  the  laws  of  §  103. 

Hence,  if  the  laws  hold  for  the  quotient  of  the  difference  of 
two  like  powers  of  a  and  b  divided  by  a  —  6,  they  also  hold  for 
the  quotient  of  the  difference  of  the  next  higher  powers  of  a 
and  b  divided  hj  a  —  b. 

•^  fj,5 7^5 

But  we  know  that  they  hold  for  ,  and  therefore  they 

a  —  b 

A|6     7j6  ,y6    7j6 

hold  for ;  and  since  they  hold  for  ,  they  hold  for 

a-b'  ^  a-b'       ^ 

:  and  so  on. 

a  —  b 


MATHEMATICAL  INDUCTION  423 

Qn Jyn. 

Hence,  the  laws  hold  for  ,  where  n  is  any  positive 

.  ,  a  —  h 

integer. 

Putting  —  h  for  h  in  (1),  we  have 

a-(-6)                         \       J-r       -r\       } 
If  n  is  even,  (-  6)"  =  6^  and  (-  6)"-^  =  -  6"-\ 
Whence,  ^"  ~  ^"  =  a^-i  -  a^'-'^h  +  a^-^d^ 6-i.  (2) 

If  n  is  odd,  (-  by  =  -  6",  and  (-  6)"-^  =  +  h''-\ 

Whence,  ^^i^±^  =  a-^  -  a^-'b  +  a^'-'b' +  6"-i.  (3) 

a-\-b 

Equations  (2)  and  (3)  are  in  accordance  with  the  laws  of 
§103. 

467.  We  will  now  prove  that  the  law  of  §  204  holds  for  the 
square  of  a  polynomial  of  any  number  of  terms. 

Assume  the  law  to  hold  for  the  square  of  a  polynomial  of  m 
terms,  where  m  is  any  positive  integer ;  that  is, 

(a-\-b  +  c-] \-l-\-my 

=  a^-^b^-{- ■"-\-m^-\-2a(b  +  c-] hm) 

4-26(c+--+m)  +  -.=  4-2Zm.  (1) 

Then,  (a  +  6  +  c+ •-   +  m  +  n)^ 

=  (a  +  b-\-c+--+my 

4-2(a-i-6  +  cH \-m)n-^n^,  by  §  97, 

=  a^  +  b-  +  c^-\ \-m^  +  7i^ 

-\-2a{b  +  c-\ hm  +  w) 

+  2b(c-\ \-m-\-n)-\ \-2mn,  by  (1). 

This  result  is  in  accordance  with  the  law  of  §  204. 


424  ALGEBRA 

Hence,  if  the  law  holds  for  the  square  of  a  polynomial  of  m 
terms,  where  m  is  any  positive  integer,  it  also  holds  for  the 
square  of  a  polynomial  of  m  + 1  terms. 

But  we  know  that  the  law  holds  for  the  square  of  a  polyno- 
mial of  three  terms,  and  therefore  it  holds  for  the  square  of  a 
polynomial  of  four  terms ;  and  since  it  holds  for  the  square  of 
a  polynomial  of  four  terms,  it  also  holds  for  the  square  of  a 
polynomial  of  five  terms  ;  and  so  on. 

Hence,  the  law  holds  for  the  square  of  any  polynomial. 

468.  As  another  illustration  of  the  method,  we  will  prove 
that  the  sum  of  the  first  n  terms  of  the  arithmetic  progression, 

a,  a  +  d,  a 4- 2c?,  •••, 
is  given  by  the  formula  na  _|_^v^^—   )  ^^     (Compare  §  361.) 
The  sum  of  the  first  two  terms  is  2a-\-d,  which  can  be 

written  in  the  form  2a-f-        ~    ^d- 

z 

Then,  the  formula  holds  for  the  sum  of  the  first  two  terms. 
Assume  that  the  formula  holds  for  the  sum  of  the  first  ?i 
terms. 

That  is,  the  sum  of  the  first  n  terms  =  na  -f    ^    ~"    ^d 

2i 

Now  the  {n  +  l)th  term  of  the  progression  is  a  +  nd. 
Whence,  the  sum  of  the  first  {n  + 1)  terms  equals 

na  +  ^^^^^^^^(^  +  a  +  nf«  =  (n  +  l)a  +  ^(n  - 1  +  2) 

=  («4-l)«  +  ^^^d. 

This  result  is  in  accordance  with  the  formula. 

Hence,  if  the  formula  holds  for  the  sum  of  the  first  n  terms, 
it  also  holds  for  the  sum  of  the  first  n-\-\  terms. 

But  we  know  that  the  formula  holds  for  the  sum  of  the  first 
two  terms,  and  hence  it  holds  for  the  sum  of  the  first  three 
terms ;  and  since  it  holds  for  the  sum  of  the  first  three  terms, 
it  also  holds  for  the  sum  of  the  first  four  terms ;  and  so  on. 


MATHEMATICAL   INDUCTION  425 

Hence,  the  formula  holds  for  the  sum  of  the  first  n  terms, 
where  n  is  any  positive  integer. 

EXERCISE  198 

1.  Prove  that  the  sum  of  the  first  n  terms  of  the  series  1,  3, 
5,  •••  is  n^. 

2.  Prove  that  the  sum  of  the  first  n  terms  of  the  series  3,  6, 
9,  ...isM!L+D. 

3.  Prove  that  the  sum  of  the  first  n  terms  of  the  series 
111  .        n 

IS 


1.2'  2-3'  3-4'  n  +  1 

4.  Prove,  by  mathematical  induction,  that  the  sum  of  the 
first  n  terms  of  the  geometric  progression, 

a,  ar,  ar^^  •••, 

is  given  by  the  formula  S  =  ^^^"^  ~  ^^  (§  370). 

r  —  1 

5.  Prove  that  the  sum  of  the  first  n  terms  of  the  series  2*^ 

42  62  ...  is  2n(n  +  l)(2n  +  l). 
'     '  3 

6.  Prove  that  the  sum  of  the  first  n  terms  of  the  series  1^ 


426  ALGEBRA 


XXXVI.  EQUIVALENT  EQUATIONS 

469.  Two  equations,  each  involving  one  or  more  unknown 
numbers,  are  said  to  be  Equivalent  when  every  solution  of  the 
first  is  a  solution  of  the  second,  and  every  solution  of  the  second 
a  solution  of  the  first. 

470.  To  solve  an  equation  involving  one  unknown  number,  Xy 
we  transform  it  into  a  series  of  equations,  which  lead  finally  to 
the  value  of  x. 

We  have  assumed,  in  passing  from  any  equation  to  any  other, 
in  this  series,  that  every  solution  of  the  first  was  a  solution  of 
the  second,  and  every  solution  of  the  second  a  solution  of  the 
first ;  so  that  it  was  legitimate  to  use  the  second  in  place  of  the 
first  to  find  the  value  of  the  unknown  number. 

That  is,  we  have  assumed  that  the  two  equations  were  equiva- 
lent (§  469). 

We  will  now  prove  some  theorems  in  regard  to  equivalent 
equations. 

471.  If  the  same  expression  he  added  to  both  members  of  an 
equation,  the  resulting  equation  will  he  equivalent  to  the  first. 

Let  A=^B  (1) 

be  an  equation  involving  one  or  more  unknown  numbers. 
To  prove  the  equation  A  +  C=B-\-C,  (2) 

where  C  is  any  expression,  equivalent  to  (1). 

Any  solution  of  (1),  when  substituted  for  the  unknown  num- 
bers, makes  A  identically  equal  to  B  (§  79). 

It  then  makes  A-{-C  identically  equal  to  jB  +  C  (§  84,  1). 

Then  it  is  a  solution  of  (2). 

Again,  any  solution  of  (2),  when  substituted  for  the  unknown 
numbers,  makes  A-\-  C  identically  equal  to  B -\-  C. 


EQUIVALENT   EQUATIONS  427 

It  then  makes  A  identically  equal  to  B  (§  84,  2). 
Then  it  is  a  solution  of  (1). 
Therefore,  (1)  and  (2)  are  equivalent. 

The  principle  of  §  84,  1,  is  a  special  case  of  the  above. 

472.  The  demonstration  of  §  471  also  proves  that 

If  the  same  expression  he  subtracted  from  both  members  of  an 
equation,  the  resulting  equation  will  be  equivalent  to  the  first. 

The  principle  of  §  84,  2,  is  a  special  case  of  this. 

473.  If  the  members  of  an  equation  be  multiplied  by  the  same 
expi'ession,  ichich  is  not  zero,  and  does  not  involve  the  unknown 
numbers,  the  resaltirig  equation  will  be  equivalent  to  the  first. 

Let  A  =  B  (1) 

be  an  equation  involving  one  or  more  unknown  numbers. 

To  prove  the  equation  AxG=Bx  C,  (2) 

where  C  is  not  zero,  and  does  not  involve  the  unknown  num- 
bers, equivalent  to  (1). 

Any  solution  of  (1),  when  substituted  for  the  unknown  num- 
bers, makes  A  identically  equal  to  B. 

It  then  makes  Ax  C  identically  equal  to  .B  x  0  (§  84,  3). 

Then  it  is  a  solution  of  (2). 

Again,  any  solution  of  (2),  when  substituted  for  the  unknown 
numbers,  makes  Ax  C  identically  equal  to  B  x  C. 

It  then  makes  A  identically  equal  to  B  (§  84,  4). 

Then  it  is  a  solution  of  (1). 

Therefore,  (1)  and  (2)  are  equivalent. 

The  reason  why  the  above  does  not  hold  for  the  multiplier  zero  is,  that 
the  principle  of  §  84,  4,  does  not  hold  when  the  divisor  is  zero. 
The  principle  of  §  84,  3,  is  a  special  case  of  the  above. 

474.  If  the  members  of  an  equation  be  multiplied  by  an  ex- 
pression which  involves  the  unknown  numbers,  the  resulting 
equation  is,  in  general,  not  equivalent  to  the  first. 

Consider,  for  example,  the  equation  a;  +  2  =  3  a?  —  4.  (1) 


428  ALGEBRA 

Now  the  equation 

(x-{.2)(x-l)  =  (Sx-^)(ix-l),  (2) 

which  is  obtained  from  (1)  by  multiplying  both  members  by 
ic  —  1,  is  satisfied  by  the  value  x  =  l,  which  does  not  satisfy  (1). 

Then  (1)  and  (2)  are  not  equivalent. 

Thus  it  is  never  allowable  to  multiply  both  members  of  an  inte- 
gral equation  by  an  expression  which  involves  the  unknown 
numbers ;  for  in  this  way  additional  solutions  are  introduced. 

475.  If  the  merribers  of  an  equation  he  divided  by  the  same  ex- 
pression, which  is  not  zero,  and  does  not  involve  the  unknown 
numbers,  the  resulting  equation  will  be  equivalent  to  the  first. 

Let  A  =  B  (1) 

be  an  equation  involving  one  or  more  unknown  numbers. 

A     B 

To  prove  the  equation        —  —  —^  (2) 

where  C  is  not  zero,  and  does  not  involve  the  unknown  num- 
bers, equivalent  to  (1). 

Any  solution  of  (1),  when  substituted  for  the  unknown  num- 
bers, makes  A  identically  equal  to  B. 

A  B 

It  then  makes  —  identically  equal  to  —  (§  84,  4). 
G  C 

Then  it  is  a  solution  of  (2). 

Again,  any  solution  of  (2),  when  substituted  for  the  unknown 

A  ■         •  B 

numbers,  makes  —  identically  equal  to  —  • 
G  G 

It  then  makes  A  identically  equal  to  B. 

Then  it  is  a  solution  of  (1). 

Therefore,  (1)  and  (2)  are  equivalent. 

The  principle  of  §  84,  4,  is  a  special  case  of  the  above. 

476.  If  the  members  of  an  equation  be  divided  by  an  ex- 
pression which  involves  the  unknown  numbers,  the  resulting 
equation  is,  in  general,  not  equivalent  to  the  first. 


EQUIVALENT   EQUATIONS  429 

Consider,  for  example,  the  equation 

(a;  +  2)(a;-l)  =  (3a;-4)(a;-l).  (1) 

Also  the  equation         x  +  2  =  Sx  —  4.,  (2) 

which  is  obtained  from  (1)  by  dividing  both  members  by  x  —  1. 

Now  equation  (1)  is  satisfied  by  the  value  x  =  l,  which  does 
not  satisfy  (2). 

Then  (1)  and  (2)  are  not  equivalent. 

It  follows  from  this  that  it  is  never  allowable  to  divide  both 
members  of  an  integral  equation  by  an  expression  which  in- 
volves the  unknown  numbers ;  for  in  this  way  solutions  are  lost. 
(Compare  §  158.) 

477.  If  both  members  of  a  fractional  equation  be  multiplied  by 
the  L.C.M.  of  the  given  denoyninators,  the  resulting  equation  is, 
in  general,  equivalent  to  the  first. 

Let  all  the  terms  be  transposed  to  the  first  member,  and  let 
them  be  added,  using  for  a  common  denominator  the  L.  C.  M. 
of  the  given  denominators. 

The  equation  will  then  be  in  the  form 

1  =  0.  (1) 

We  will  now  prove  the  equation 

A  =  0,  (2) 

which  is  obtained  by  multiplying  (1)  by  the  L.  C.  M.  of  the 
given  denominators,  equivalent  to  (1),  if  A  and  B  have  no  com- 
mon factor. 

Any  solution  of  (1),  when  substituted  for  the  unknown  num- 
bers, makes  —  identically  equal  to  0. 

Then,  it  must  make  A  identically  equal  to  0. 
Then,  it  is  a  solution  of  (2). 

Again,  any  solution  of  (2),  when  substituted  for  the  unknown 
numbers,  makes  A  identically  equal  to  0. 


430  ALGEBRA 

Since  A  and  B  have  no  common  factor,  B  cannot  be  0  when 
this  solution  is  substituted  for  the  unknown  numbers. 

Then,  any  solution  of  (2),  when  substituted  for  the  unknown 

numbers,  makes  —  identically  equal  to  0,  and  is  a  solution  of  (1). 
B 

Therefore,  (1)  and  (2)  are  equivalent,  if  A  and  B  have  no 

common  factor. 

If  A  and  B  have  a  common  factor,  (1)  and  (2)  are  not  equivalent; 
consider,  for  example,  the  equations 

^^^^  =  0,  and  X  -  1  =  0. 

The  second  equation  is  satisfied  by  the  value  x  =  1,  which  does  not 
satisfy  the  first  equation  ;  then,  the  equations  are  not  equivalent. 

478.  A  fractional  equation  may  be  cleared  of  fractions  by 
multiplying  both  members  by  any  common  multiple  of  the 
denominators;  but  in  this  way  additional  solutions  are  intro- 
duced, and  the  resulting  equation  is  not  equivalent  to  the  first. 

Consider,  for  example,  the  equation 


-1 


If  we  solve  by  multiplying  both  members  by  x^—1,  the 
L.  C.  M.  of  cc^  —  1  and  a^  —  1,  we  find  x  =  —  2. 

If,  however,  we  multiply  both  members  by  (x^  —  l){x  —  1), 
we  have 

a^-oc^-{-a^-x=:2x^-2x^-2x-^2,  or  a^-{-x-2  =  0. 

The  latter  equation  may  be  solved  as  in  §  126. 
The  factors  oi  xr  -\-  x  ~2  sue  x-\-2  and  x  —  1. 
Solving  the  equation  .'^^  +  2  =  0,  x=  —  2. 
Solving  the  equation  .t  —  1  =  0,  a?  =  1. 

This  gives  the  additional  value  x=l;  and  it  is  evident  that 
this  does  not  satisfy  the  given  equation. 

479.  If  both  members  of  an  equation  be  raised  to  the  same 
positive  integral  power,  the  resulting  equation  ivill  have  all  the 
solutions  of  the  given  equation,  and,  in  general,  additional  ones. 


EQUIVALENT  EQUATIONS  431 

Consider,  for  example,  the  equation  x  =  3. 
Squaring  both  members,  we  have 

aj2  =  9,  or  a;--9  =  0,  or  (x-{-3){x-S)  =  0. 

The  latter  equation  has  the  root  3,  and,  in  addition,  the 
root  —  3. 

We  will  now  consider  the  general  case. 

Let  A  =  B  (1) 

be  an  equation  involving  one  or  more  unknown  numbers. 

Raising  both  members  to  the  nth  power,  n  being  a  positive 
integer,  we  have 

A^'  =  B-,  or  A-  -  B''  =  0.  (2) 

Factoring  the  first  number  (§  121), 

(A  -  B)  (A""-'  +  A^-'B  +  . . .  +  5«"i)  =  0.  (3) 

Now,  equation  (3)  is  satisfied  when  A=B. 
Whence,  equation  (2)  has  all  the  solutions  of  (1). 
But  (3)  is  also  satisfied  when 

so  that  (2)  has  also  the  solutions  of  this  last  equation,  which, 
in  general,  do  not  satisfy  (1). 

EQUIVALENT  SYSTEMS  OF  EQUATIONS 

480.  Two  systems  of  equations,  involving  two  or  more 
unknown  numbers,  are  said  to  be  equivalent  when  every  solu- 
tion of  the  first  system  is  a  solution  of  the  second,  and  every 
solution  of  the  second  a  solution  of  the  first. 


481.  // 


are   equations   involving  tioo   or  more   unknown  numbers,   the 
system  of  equations 

A=0, 
mA  +  nB  =  0, 

where  m  and  n  are  any  numbers,  and  n  not  equal   to  zero,   is 
equivalent  to  the  Jirst  system. 


432  ALGEBRA 

For  any  solution  of  the  first  system,  when  substituted  for 
the  unknown  numbers,  makes  A  =  0  and  ^  =  0. 

It  then  makes  ^  =  0  and  mA  +  nB  =  0. 

Then,  it  is  a  solution  of  the  second  system. 

Again,  any  solution  of  the  second  system,  when  substituted 
for  the  unknown  numbers,  makes  ^  =  0  and  mA  +  nB  =  0. 

It  therefore  makes  nB  =  0,  oi  B  =  0. 

Since  it  makes  A  =  0  and  5  =  0,  it  is  a  solution  of  the  first 
system. 

Hence,  the  systems  are  equivalent. 

A  similar  result  holds  for  a  system  of  any  number  of  equations. 
Either  m  or  n  may  be  negative. 

482.  If  either  equation,  in  a  system  of  tvjo,  be  solved  for  one 
of  the  unknown  numbers,  and  the  value  found  be  substituted  for 
this  unknown  number  in  the  other  equation,  the  resulting  system 
will  be  equivalent  to  the  first. 

Let  H  =  ^^  W 

la=A  (2) 

be  equations  involving  two  unknown  numbers,  x  and  y. 

Let  E  be  the  value  of  x  obtained  by  solving  (1). 

Let  F=G  be  the  equation  obtained  by  substituting  E  for 
X  in  (2). 

To  prove  the  system  of  equations 

'x=E,  '  (3) 

\f=G,  (4) 

equivalent  to  the  first  system. 

Any  solution  of  the  first  system  satisfies  (3),  for  (3)  is  only 
a  form  of  (1). 

Also,  the  values  of  x  and  y  which  form  the  solution  make  x 
and  E  equal ;  and  hence  satisfy  the  equation  obtained  by  put- 
ting E  for  x  in  (2). 

Then,  any  solution  of  the  first  system  satisfies  (4). 

Again,  any  solution  of  the  second  system  satisfies  (1),  for 
(1)  is  only  a  form  of  (3). 


EQUIVALENT  EQUATIONS  433 

Also,  the  values  of  x  and  y  which  form  the  solution  make  x 
and  E  equal ;  and  hence  satisfy  the  equation  obtained  by  put- 
ting X  for  E  in  (4). 

Then,  any  solution  of  the  second  system  satisfies  (2). 

Hence,  the  systems  are  equivalent. 

A  similar  result  holds  for  a  system  of  any  number  of  equations, 
involving  any  number  of  unknown  numbers. 

483.  We  will  now  apply  the  principles  of  §  §  481  and  482 
to  show  that  the  solutions  of  Ex.  1,  §  168,  and  the  examples  of 
§§  169  and  170  are  equivalent  to  the  given  equations.  ' 

Ex.  1,  §  168. 

By  §  481,  the  given  system  is  equivalent  to  the  system  (1) 
and  (5),  or  to  the  system  (1)  and  (6). 

By  §  482,  the  system  (1)  and  (6)  is  equivalent  to  the  system 
(6)  and  (7),  which  is  equivalent  to  the  system  (6)  and  (8). 

Then,  the  given  system  is  equivalent  to  the  system  (6)  and  (8). 

Ex.,  §  169. 

By  §  482,  the  given  system  is  equivalent  to  the  system  (3) 
and  (4),  or  to  the  system  (3)  and  (5). 

By  §  482,  the  system  (3)  and  (5)  is  equivalent  to  (5)  and  (6). 

Ex.,  §  170. 

The  given  system  is  equivalent  to  (3)  and  (4). 

Now  any  values  of  x  and  y  which  satisfy  (3)  and  (4)  also 
satisfy  (3)  and  (5). 

Then,  the  given  system  is  equivalent  to  the  system  (3)  and 
(5),  or  to  (3)  and  (6). 

By*§  482,  the  system  (3)  and  (6)  is  equivalent  to  (6)  and  (7). 

484.  The  principles  of  §§  471,  472,  473,  475,  477,  479,  480, 
and  481  hold  for  equations  of  any  degree. 


434  ALGEBRA 


XXXVII.    GRAPHICAL  REPRESENTATION  OF 
IMAGINARY  NUMBERS 

485.   Let  0  be  any  point  in  the  straight  line  XX'. 
We   may    suppose   any  positive    real    number,    -f-  a,   to    be 
represented   by  the   distance   from 


O  to  ^,  a  units  to  the  right  of  0     „>      .      ^     .     ^        .     ^ 
in  OX,  ^     A'   -a    0  -.a     A       ' 

Then,  with  the  notation  of  §  16,  any  negative  real  number, 
—  a,  may  be  represented  by  the  distance  from  0  to  A^  a  units 
to  the  left  of  0  in  OX. 

486.  Since  —  a  is  the  same  as  (+  a)  x  (—  1),  it  follows  from 
§  485  that  the  product  of  +  a  by  —  1  is  represented  by  turning 
the  line  OA  which  represents  the  number  -f  a,  through  two 
right  angles,  in  a  direction  opposite  to  the  motion  of  the  hands 
of  a  clock. 

Then,  in  the  product  of  any  real  number  by  —1,  we  may 
regard  —  1  as  an  operator  which  turns  the  line  which  repre- 
sents the  first  factor  through  two  right  angles,  in  a  direction 
opposite  to  the  motion  of  the  hands  of  a  clock. 

487.  Graphical  Representation  of  the  Imaginary  Unit  i  (§  276). 
By  the  definition  of  §  275,  —  1  =  i  x  i. 
Then,    since    one    multiplication    by   z,  „ 

followed  by  another  multiplication  by  *, 

turns  the  line  which  represents  the  first  +z 

factor  through  two  right  angles,  in  a  direc- 


tion opposite  to  the  hands  of  a  clock,  we  -i 

may  regard  multiplication  by  i  as  turning  '' 

the  line  through  one  right  angle,  in  the 
same  direction.  ^ 

Thus,  let  XX  and   YY'  be  straight  lines  intersecting  at 
right  angles  at  0. 


^ai 


^ 


0-V(^   A. 
-ai 


REPRESENTATION  OF   IMAGINARY  NUMBERS    435 

Then,  if  +  a  be  represented  by  the  line  OA,  where  A  is  a 
units  to  the  right  of  0  in  OX,  +ai  may  be  represented  by 
OB,  and  —  ai  by  OB',  where  ^  is  a  units  above,  and  B'  a 
units  below,  0,  in  YY'. 

Also,  H-  i  may  be  represented  by  OC,  and  —  i  by  0(7',  where 
C  is  one  unit  above,  and  C  one  unit  below,  0,  in  YY'. 

488.   Graphical  Representation  of  Complex  Numbers. 

We  will  now  show  how  to  represent  the  complex  number 
a  +  bi. 

Let  XX'  and  YY'  be  straight  lines  inter- 
secting at  right  angles  at  0. 

Let  a  be  represented  by  OA,  to  the  right    ^_ 
of  0,  if  a  is  positive,  to  the  left  if  a  is 
negative. 

Let  bi  be  represented  by  OB,  above  0  if  6  is  positive,  below 
if  b  is  negative. 

Draw  line  AC  equal  and  parallel  to  OB,  on  the  same  side  of 
XX'  as  OB,  and  line  OC. 

Then,  OC  is  considered  as  representing  the  result  of  adding 
bito  a;  that  is,  OC  represents  the  complex  number  a  +  bi. 

The  figure  represents  the  case  in  which  both  a  and  b  are  positive. 

As  another  illustration,  we  will  show  how  to  represent  the 
complex  number  —  5  —  4  i 

Lay  off  OA  5  units  to  the  left  of  0  in 
OX',  and  OB  4  units  below  0  in  YY'. 

Draw  line  AC  below  XX',  equal  and 
parallel  to  OB,  and  line  OC. 

Then,  00  represents  —  5  —  4  i. 

The  complex  number  a  +  bi,  if  «  is  positive  and  b  negative,  will  be 
represented  by  a  line  between  OX  and  0  Y' ;  and  if  a  is  negative  and  b 
positive,  by  a  line  between  0  Y  and  OX'. 

EXERCISE  199 

Represent  the  following  graphically : 

1.   3i.  2.   -6i.  3.   4  +  i.  4.    -l  +  2i. 


\''-J-     ~^ 

0 

X 

y 

1 

c 

J 

<s>. 

B 

y' 

436 


ALGEBRA 


5.    2-5i. 


6.    -5-3i. 


7.    -7  +  4i. 


489.   Graphical  Representation  of  Addition. 

We  will  now  show  how  to  represent  the  result  of  adding  b  to 
a,  where  a  and  b  are  any  two  real,  pure  imaginary,  or  complex 
numbers. 

Let  the  line  a  be  represented  by  OA,  and 
the  line  b  by  OB. 

Draw  the  line  AC  equal  and  parallel  to 
OB,  on  the  same  side  of  OA  as  OB,  and  the 
line  00. 

Then,  OG  is  considered  as  representing 
the  result  of  adding  b  to  a;  that  is,  00  represents  a-\-b. 

The  method  of  §  488  is  a  special  case  of  the  above. 

If  a  and  6  are  both  real,  B  will  fall  in  OA,  or  in  AO  produced 
through  0. 

The  same  will  be  true  if  a  and  6  are  both  pure  imaginary. 

If  one  of  the  numbers,  a  and  &,  is  real,  and  the  other  pure  imaginary, 
the  lines  OA  and  OB  will  be  perpendicular. 

As  another  illustration,  we  will  show  how  to  represent 
graphically  the  sum  of  the  complex  numbers  2  —  5 1  and 
-4  +  3i. 

The  complex  number  2  —  5i  is  repre- 
sented by  the  line  OA,  between  OX  and 
OY'. 

The  complex  number  —  4  -f-  3  if  is  repre- 
sented by  the  line  OB,  between  OT  and 
OX'. 

Draw  the  line  BC  equal  and  parallel  to 
OA,  on  the  same  side  of  OB  as  OA,  and  the  line  OC. 

Then,  the  line  00  represents  the  result  of  adding 
to  2  -  5 1. 


4-1-3/ 


490.  Graphical  Representation  of  Subtraction. 

Let  a  and  b  be  any  two  real,  pure  imaginary,  or  complex 
numbers. 


REPRESENTATION  OF   IMAGINARY  NUMBERS      437 


Let  a  be  represented  by  OA,  and  b 
by  OB;  and  complete  the  parallelogram 
OBAC. 

By  §  489,  OA  represents  the  result  of 
adding  the  number  represented  by  OB  to 
the  number  represented  by  00. 

That  is,  if  b  be  added  to  the  number 
represented  by  0(7,  the  sum  is  equal  to 
a;   hence,  a  — 6  is  represented  by  the  line  OC. 

EXERCISE  200 
Represent  the  following  graphically : 

1.  The  sum  of  4  i  and  3  —  5i. 

2.  The  sum  of  —  5  ^  and  —  1  +  6  i. 

3.  The  sum  of  2  +  4  i  and  5  -  3  i. 

4.  The  sum  of  -6  +  2i  and  -4-7z. 

5.  Represent  graphically  the  result  of  subtracting  the   sec- 
ond expression  from  the  first,  in  each  of  the  above  examples. 


438  ALGEBRA 


XXXVIII.    INDETERMINATE  FORMS 

491.  In  §  322,  we  found  that  the  form  -  indicated  an  ex- 
pression which  could  have  any  value  whatever ;  but  this  is  not 
always  the  case. 

^  rjf^  rt^  ' 

Consider,  for  example,  the  fraction  — . 

or —  ax 

If  x  =  a,  the  fraction  takes  the  form  -. 

JSTow  x^-a^  ^(x-\-a)(x-a)^x-{-a^ 

x^—  ax  x(x  —  a)  x    ^ 

which  last  expression  is  equal  to  the  given  fraction,  provided 
X  does  not  equal  a. 

The  fraction  ^_ZL^  approaches  the  limit  ^L+_^    or  2,  when 
X  a 

X  approaches  the  limit  a. 

This  limit  we  call  the  value  of  the  given  fraction  ivhen  x  =  a. 

Then,  the  value  of  the  given  fraction  when  x  =  a  h  2. 

In  any  similar  case,  we  cancel  the  factor  which  equals  0  for  the  given 
value  of  X,  and  find  the  limit  approached  by  the  result  when  x  approaches 
the  given  value  as  a  limit. 

EXERCISE  201 
Find  the  values  of  the  following : 

1.   ^«^-^^'when^  =  2a.         3.    -/-^li_  when  a;  =  - 4. 
x'-4:a'  x'-\-2x-S 

'•    4^T3^  when  .  =  0.  4.   g^,_^,^^,,  when  .  =  i. 

g_  ^dh6^±12^±8^hen.:  =  -2. 
x*-8  X-  + 16 

6.   ^l^M±l£^whenx  =  2. 
x'-Tx  +  G 


INDETERMINATE   FORMS  439 


492.   Other  Indeterminate  Forms. 

Expressions  taking  the  forms  ^, 
values  of  the  letters  involved,  are  also  indeterminate. 


Expressions  taking  the  forms  ^,  0  x  oo,  or  oo  —  oo,  for  certain 


1.   Find  the  value  of  (o(^ -\- 8)(l  + -^-^  when  x  =  -2. 
This  expression  takes  the  form  0  x  oo,  when  x  =  —  2  (§  319). 

Now,  (x3  +  8) (l  +  —^)  =  x3  +  8  •  ^^  +  ^ 
\        X  +  2/ 


'^x  +  2 
=  x3  +  8  4-  ic2  -  2  X  +  4  =  x3  +  x2  -  2  X  +  12. 

The  latter  expression  approaches  the  limit  —8  +  4  +  4  +  12,  or  12, 
when  X  approaches  the  limit  —  2. 

This  limit  we  call  the  value  of  the  expression  when  x  =  —  2  ;  then,  the 
value  of  the  expression  when  x  =  —  2,  is  12. 

In  any  similar  case,  we  simplify  as  much  as  possible  before  finding  the 
limit. 

2.   Find  the  value  of  -^ ?-^,  when  x==l. 

1  —  X     1  —  or 

The  expression  takes  the  form  oo  —  oo,  when  x  =  1  (§  319). 

2x        1+X-2X      1-x         1 


Now, 


1-x        1-X2  1-X2  1-X2        1+X 


The  latter  expression  approaches  the  limit  -  when  x  approaches  the 
limit  1.  ^ 

Then,  the  value  of  the  expression  when  x  =  1,  is  -  • 

493.   Another  example  in  which  the  result  is  indeterminate 
is  the  following : 

Ex.   Find  the  limit  approached  by  the  fraction      "^  ^  when 
X  is  indefinitely  increased.  ~     ^ 

Both  numerator  and  denominator  increase  indefinitely  in  absolute  value 
when  X  is  indefinitely  increased. 

Dividing  each  term  of  the  fraction  by  x,  -^ — -  = 

^  "^'2-5x2^ 

—  o 

X 


440  ALGEBRA 


The  latter  expression  approaches  the  limit  ^  (§  320),  or  — -,  when 
X  is  indefinitely  increased.  ~  '^ 

In  any  similar  case,  we  divide  both  numerator  and  denominator  of  the 
fraction  by  the  highest  power  of  x. 


EXERCISE   202 

Find  the  limits  approached  by  the  following  when  x  is  in- 
definitely increased: 

1    4  +  5a;-3^  2     ^^  +  1  3    ^-2x-^ 

7  -  a;  4-  4  0)2  '  '    3  aj^  -  2*  '   x^-\-bx^3 


Find  the  values  of  the  following : 

1  12 

a;_2     aj3-8 


4.  r when  x  =  2. 


5.    (2a^-5a;-3)/'2-f-^')  whena;  =  3. 


HORNER'S  SYNTHETIC   DIVISION 


441 


XXXIX.    HORNER'S  SYNTHETIC  DIVISION 

494.   Division  by  Detached  Coefficients. 

In  finding  the  quotient  of  two  expressions  which  are  arranged 
according  to  the  same  order  of  powers  of  some  common  letter, 
the  operation  may  be  abridged  by  writing  only  the  jiumerical 
coefficients  of  the  terms. 

If  the  term  involving  any  power  is  wanting,  it  may  be  sup- 
plied with  the  coefficient  0. 

Ex.   Divide  6  x^-\-2  a^-9  a;^+5  a^  + 18  i»-30  by  3  a^  +  a^-6. 

3  +  l4-0_6 
2+0-3+5 


6  +  2- 

-9+   0  +  5  +  18- 

fO-12 

-9  +  12 

_9-   3  +  0  +  18 

-30 

6  +  2- 

15  +  5 
15-j_5-f  0- 

-30 

Then  the  quotient  is  2  a^  —  3  a;  +  5. 


495.  Horner's  Synthetic  Division. 

Let  it  be  required  to  divide  6  x^  —  ic^  —  3  a^  + 10  a?  — 12  by 
2a^  +  x-S. 


ex*-    a^-3a^  +  10a; 
6x*-\-3a^-9x^ 


12 


-4ar^  +  6a;2 
-4a^-2a^  + 

ex 

Sx'  + 
8a.'2  + 

Ax 
Ax- 

-12 

2a^  +  a?-3 


3a^_2a;  +  4 


The  dividend  equals  (2  a;^  +  x  —  3)  times  the  quotient. 

Then,  we  can  find  the  quotient  by  subtracting  from  the 
dividend  +  x  times  the  quotient,  and  —  3  times  the  quotient, 
and  dividing  the  result  by  2  a;^. 


442  ALGEBRA 

Or,  we  can  find  it  by  adding  to  the  dividend  —x  times  the 
quotient,  and  +  3  times  the  quotient,  and  dividing  the  result 
by2a^. 

We  may  arrange  the  work  as  follows  : 


2a^ 

60)4-     a?-Zx'-\-10x-12 

—  X 

-3a^  +  2aj2-    4^ 

+  3 

+  9a^-    Qx  +  12 

^x'-2x  4-4,  quo.     0      0 

We  write  the  divisor  in  the  left-hand  column,  with  the  sign 
of  each  term  after  the  first  changed. 

We  get  the  first  term  of  the  quotient,  3  a^,  by  dividing  the  first 
term  of  the  dividend,  6  a?'^,  by  the  first  term  of  the  divisor,  2  x^. 

Since  we  have  to  add  to  the  dividend  —  x  times  the  quotient, 
and  +3  times  it,  we  can  put  down  the  terms  —  3a^  in  the 
second  column,  and  4-  9  x"  in  the  third,  these  being  the  prod- 
ucts of  —  a;  and  +  3  by  the  first  term  of  the  quotient. 

Now  we  get  the  second  term  of  the  quotient  by  subtracting 
3  ic^  from  —  a^,  and  dividing  the  result  by  2  x^. 

Then,  in  the  synthetic  method,  we  add  the  terms  in  the 
second  column,  and  divide  the  sum  by  2  x^,  giving  —  2  a?. 

We  now  write  the  term  +  2  a^  in  the  second  column,  and 
—  6a;  in  the  third;  these  being  the  products  of  —a;  and  -|-3 
by  the  second  term  of  the  quotient. 

We  get  the  third  term  of  the  quotient  by  subtracting  —  9  a^ 
and  —  2  a;^  from  —  3  x^,  and  dividing  the  result  by  2  x?. 

Then,  in  the  synthetic  method,  we  add  the  terms  in  the 
third  column,  and  divide  the  sum  by  2  oi?,  giving  4. 

We  now  write  the  term  —  4  a;  in  the  fourth  column,  and  + 12 
in  the  fifth  ;  these  being  the  products  of  —  a;  and  +  3  by  the 
third  term  of  the  quotient. 

We  have  now  added  to  the  dividend  —  x  times  the  quotient, 
and  +  3  times  the  quotient,  and  divided  the  result  by  2  a^ ;  so 
that  we  have  obtained  the  quotient. 

The  sum  of  the  terms  in  the  fourth  and  fifth  columns  is  0  ;  if  this  had 
not  been  the  case,  there  would  have  been  a  remainder. 


HORNER'S  SYNTHETIC   DIVISION  443 

496.   We  will  now  give  some  additional  examples  of  the 
method : 

1.   Divide         12  ar^ -11  aj^-f  20  a;- 9  by  3  ar-2a;  +  4. 


3x2 
-4 


12  a:3  -  11  a;2  +  20  X  -  9 
+    8x2--  2  X 

-16x  +  4 


4  X  —  1,  quo.    2  X  —  5,  Rem. 


"We  write  the  divisor  in  the  left-hand  column,  with  the  sign  of  each 
term  after  the  first  changed. 

Dividing  12  x^  by  3  x^  gives  4  x  for  the  first  term  of  the  quotient. 

We  multiply  +  2  x  by  4  x  and  put  the  product,  8  x2,  in  the  second 
column  ;  and  multiply  —  4  by  4  x,  and  put  the  product,  —  16  x,  in  the 
third  column. 

We  add  the  terms  in  the  second  column,  giving  —  3  x2,  and  divide  the 
result  by  3  x2,  giving  —  1  as  the  second  term  of  the  quotient. 

We  multiply  +  2  x  by  —  1  and  put  the  product,  —  2  x,  in  the  third 
column  ;  and  multiply  —  4  by  —  1,  and  put  the  product,  +  4,  in  the 
fourth  column. 

Adding  the  terms  in  the  third  and  fourth  columns,  the  sum  is  2  x  —  5. 

Then,  the  quotient  is  4  x  —  1,  and  the  remainder  2  x  —  5. 

It  is  advantageous  to  use  detached  coefficients  (§  494)  in  the  synthetic 
method ;  the  work  of  Ex.  1  would  then  stand  as  follows  : 

.  3 

+  2 

-4  ^ 

4-1,     +2-5 
2.   Divide 

a«+2a^6-14a362_|_l5a6*-5  6^  by  a''-^ah-\-h\ 


12 

— 

11  4.  20  -  9 

+ 

8-    2 
-16  +  4 

+  3a6 

-&2 


^5  4.  2  a*&  -  14  aW  +  15  ah^  -  5  65 

+  3a46  +  15a362  - 15  a&* 

-      am-ba%^  +5  66 


a3  +  5a2&+   0         -5  63,         0 


In  the  above  solution,  the  sum  of  the  terms  in  the  third  column  is  0,  so 
that  the  third  term  of  the  quotient  is  0. 

We  then  add  the  terms  in  the  fourth  column  and  divide  by  a^^  giving 
—  5  6^  as  the  fourth  term  of  the  quotient. 

It  is  important,  in  an  example  like  the  above,  to  keep  similar  terms  in 
the  same  vertical  column. 


444 


ALGEBEA 


The  work  of  Ex.  2  will  appear  as  follows  with  detached  coefficients 


1 
+  3 
-1 


1  +  2-14  +  0  +  15-5 
+  3_|.15        _15 

-    1-5  +5 


1  +  5+0-5,      0     0 


EXERCISE   203 

Divide  the  following  by  synthetic  division : 

1.  12x'-7a^-23x-S  hj  4.x'-5x-3. 

2.  4a*-9a2+-30a-25  by  2a2+-3a-5. 

3.  2a'-a^b-\-Sab^-5b^  by  2a^-3ab  +  5b\ 

4.  4  m^n*  +-  n^  + 16  m'*  by  2  mn^  +-  4  m^  -f-  7i*. 

'5.   6a^-13x'-20a^  +  55a^-l4:x-19  by  2a^-7a;  +  6. 

6.  Sx^-4:x'y-SxY-18xy^-{-21y' 

by  4  a^  —  2  0^2/  +-  6  a;^^  —  7  2/"'. 

7.  37a2  +  50+-a^-70a  by  2a2+-5  +  a3-6a. 

8.  2a'-ab-2ac-6b'+llbc-4.c'  by  2a  +  36-4c. 


PERMUTATIONS  AND   COMBINATIONS  445 


XL.   PERMUTATIONS  AND   COMBINATIONS 

497.  The  different  orders  in  which  things  can  be  arranged 
are  called  their  Permutations. 

Thus,  the  permutations  of  the  letters  a,  6,  c,  taken  two  at  a 
time,  are  ahy  ac,  ha,  he,  ca,  ch ;  and  their  permutations,  taken 
three  at  a  time,  are  abc,  ach,  hac,  hca,  cah,  cha. 

498.  The  Combinations  of  things  are  the  different  collections 
which  can  be  formed  from  them  without  regard  to  the  order 
in  which  they  are  placed. 

Thus,  the  combinations  of  the  letters  a,  h,  c,  taken  two  at  a 
time,  are  ap,  he,  ca;  for  though  ah  and  ha  are  different  permu- 
tations, they  form  the  same  combination. 

499.  To  find  the  number  of  permutations  of  n  different  things 
taken  two  at  a  time. 

Consider  the  n  letters  a,  h,  c,  •••. 

In  making  any  particular  permutation  of  two  letters,  the 
first  letter  may  be  any  one  of  the  n;  that  is,  the  first  place 
can  be  filled  in  n  different  ways. 

After  the  first  place  has  been  filled,  the  second  place  can  be 
filled  with  any  one  of  the  remaining  n  —  1  letters. 

Then,  the  whole  number  of  permutations  of  the  letters  taken 
two  at  a  time  is  n(n  —  1). 

We  will  now  consider  the  general  case. 

500.  To  find  the  numher  of  permutations  of  n  different  things 
takeri  r  at  a  time. 

Consider  the  n  letters  a,  h,  c,  •••. 

In  making  any  particular  permutation  of  r  letters,  the  first 
letter  may  be  any  one  of  the  n. 

After  the  first  place  has  been  filled,  the  second  place  can  be 
filled  with  any  one  of  the  remaining  n  —  1  letters. 


446  ALGEBRA 

After  the  second  place  has  been  filled,  the  third  place  can  be 
filled  in  M  —  2  different  ways. 

Continuing  in  this  way,  the  ?*th  place  can  be  filled  in 

n  —  (r  —  l),  or  n  —  r-\-l  different  ways. 

Then,  the  whole,  number  of  permutations  of  the  letters  taken 
r  Sit  a  time  is  given  by  the  formula 

,P,  =  n(7i -  l)(n  -  2)  ...  (n -  r  + 1).  (1) 

The  number  of  permutations  of  n  different  things  taken  r  at  a  time  is 
usually  denoted  by  the  symbol  nPr- 

501.  -If  all  the  letters  are  taken,  r  =  n,  and  (1)  becomes 

.P„  =  n(n-l)(n-2)...3.2.1=[n.  (2) 

Hence,  the  number  of  permutations  of  n  different  things  taken 
n  at  a  time  equals  the  product  of  the  natural  numbers  from  1  to  n 
inclusive.     (See  note,  page  352.) 

502.  To  find  the  number  of  combinations  of  n  different  things 
taken  r  at  a  time. 

The  number  of  permutations  of  n  different  things  taken  r  at 
a  time  is       ^ ^^ ._  i)  (^  _  2)  ...  (^  _  r  + 1)  (§  500). 

But,  by  §  501,  each  combination  of  r  different  things  may 
have  \r  permutation?. 

Hence,  the  number  of  combinations  of  n  different  things  taken 
r  at  a  time  equals  the  number  of  permutations  divided  by  [r. 

That  is,      „fi  =  "('^-l)("-2)-(«-r  +  l).  ^g^ 

[r 

The  number  of  combinations  of  n  different  things  taken  r  at  a  time  is 
usually  denoted  by  the  symbol  nCv 

503.  Multiplying  both  terms  of  the  fraction  (3)  by  the  prod- 
uct of  the  natural  numbers  from  1  to  n  —  r  inclusive,  we  have 

^  ^n(n-l)...(n-r  +  l)-(^-r)...2.1^      1^      . 
[rxl.2...(n-r)  \r\n-r 

which  is  another  form  of  the  result. 


PERMUTATIONS  AND   COMBINATIONS  447 

504.  The  number  of  combinations  of  n  different  things  taken  r 
at  a  time  equals  the  number  of  combinations  taken  n  —  r  at  a  time. 

For,  for  every  selection  of  r  things  out  of  n,  we  leave  a  selec- 
tion oi  n  —  r  things. 

The  theorem  may  also  be  proved  by  substituting  n  —  r  for  r,  in  the 
result  of  §  503. 

505.  Examples. 

1.  How  many  changes  can  be  rung  with  10  bells,  taking  7  at 
a  time  ? 

Puttmg  n  =  10,  r  =  7,  in  (1),  §  500, 

10P7  =  10. 9. 8. 7. 6. 5. 4  =  604800. 

2.  How  many  different  combinations  can  be  formed  with  16 
letters,  taking  12  at  a  time  ? 

By  §  504,  the  number  of  combinations  of  16  different  things,  taken  12 
at  a  time,  equals  the  number  of  combinations  of  16  different  things,  taken 
4  at  a  time. 

Putting  n  =  16,  r  =  4,  in  (3),  §  502, 

r  _16-15-14-13_.,Qo^ 
''^'-     1.2.3.4     -^^^^* 

3.  How  many  different  words,  each  consisting  of  4  consonants 
and  2  vowels,  can  be  formed  from  8  consonants  and  4  vowels  ? 

The  number  of  combinations  of  the  8  consonants,  taken  4  at  a  time,  is 

«ll^il^,or70. 
1.2.3.4 

The  number  of  combinations  of  the  4  vowels,  taken  2  at  a  time,  is 

,  or  6. 

1.2' 

Any  one  of  the  70  sets  of  consonants  may  be  associated  with  any  one 
of  the  6  sets  of  vowels  ;  hence,  there  are  in  all  70  x  6,  or  420  sets,  each 
containing  4  consonants  and  2  vowels. 

But  each  set  of  6  letters  may  have  [6,  or  720  different  permutations 
(§  501). 

Therefore,  the  whole  number  of  different  words  is 

420  X  720,  or  302400. 


448  ALGEBRA 

EXERCISE    204 

1.  How  many  different  permutations  can  be  formed  with 
14  letters,  taken  6  at  a  time  ? 

2.  In  how  many  different  orders  can  the  letters  in  the  word 
triangle  be  written,  taken  all  together  ? 

3.  How  many  combinations  can  be  formed  with  15  things, 
taken  5  at  a  time  ? 

4.  A  certain  play  has  5  parts,  to  be  taken  by  a  company  of 
12  persons.     In  how  many  different  ways  can  they  be  assigned  ? 

5.  How  many  combinations  can  be  formed  with  17  things, 
taken  11  at  a  time  ? 

6.  How  many  different  numbers,  of  6  different  figures  each, 
can  be  formed  from  the  digits  1,  2,  3,  4,  5,  6,  7,  8,  9,  if  each 
number  begins  with  1,  and  ends  with  9  ? 

7.  How  many  even  numbers,  of  5  different  figures  each,  can 
be  formed  from  the  digits  4,  5,  6,  7,  8  ? 

8.  How  many  different  words,  of  8  different  letters  each, 
can  be  formed  from  the  letters  in  the  word  ploughed,  if  the 
third  letter  is  o,  the  fourth  u,  and  the  seventh  e  ? 

9.  How  many  different  committees,  of  8  persons  each,  can 
be  formed  from  a  corporation  of  14  persons  ?  In  how  many 
will  any  particular  individual  be  found  ? 

10.  There  are  11  points  in  a  plane,  no  3  in  the  same  straight 
line.  How  many  different  quadrilaterals  can  be  formed,  having 
4  of  the  points  for  vertices  ? 

11.  Erom  a  pack  of  52  cards,  how  many  different  hands  of 
6  cards  each  can  be  dealt  ? 

12.  A  and  B  are  in  a  company  of  48  men.  If  the  company 
is  divided  into  equal  squads  of  6,  in  how  many  of  them  will  A 
and  B  be  in  the  same  squad  ? 

13.  How  many  different  words,  each  having  5  consonants 
and  1  vowel,  can  be  formed  from  13  consonants  and  4  vowels  ? 


PERMUTATIONS  AND   COMBINATIONS  449 

14.  Out  of  10  soldiers  and  15  sailors,  how  many  different 
parties  can  be  formed,  each  consisting  of  3  soldiers  and  3 
sailors  ? 

15.  A  man  has  22  friends,  of  whom  14  are  males.  In  how 
many  ways  can  he  invite  16  guests  from  them,  so  that  10  may 
be  males  ? 

16.  From  3  sergeants,  8  corporals,  and  16  privates,  how  many 
different  parties  can  be  formed,  each  consisting  of  1  sergeant, 
2  corporals,  and  5  privates  ? 

17.  Out  of  3  capitals,  6  consonants,  and  4  vowels,  how  many 
different  words  of  6  letters  each  can  be  formed,  each  beginning 
with  a  capital,  and  having  3  consonants  and  2  vowels  ? 

18.  How  many  different  words  of  8  letters  each  can  be 
formed  from  8  letters,  if  4  of  the  letters  cannot  be  separated  ? 
How  many  if  these  4  can  only  be  in  one  order  ? 

19.  How  many  different  numbers,  of  7  figures  each,  can  be 
formed  from  the  digits  1,  2,"~^  4,  5,  6,  7,  8,  9,  if  the  first, 
fourth,  and  last  digits  are  odd  numbers  ? 

506.  To  find  the  number  of  permutations  ofn  things  which  are 
not  all  different,  taken  all  together. 

Let  there  be  n  letters,  of  which  p  are  a's,  q  are  6's,  and  r  are 
c's,  the  rest  being  all  different. 

Let  N  denote  the  number  of  permutations  of  these  letters 
taken  all  together. 

S appose  that,  in  any  particular  permutation  of  the  n  letters, 
the  p  a's  were  replaced  by  p  new  letters,  differing  from  each 
other  and  also  from  the  remaining  letters. 

Then,  by  simply  altering  the  order  of  these  p  letters  among 
themselves,  without  changing  the  positions  of  any  of  the  other 
letters,  we  could  from  the  original  permutation  form  [p  differ- 
ent permutations  (§  501). 

If  this  were  done  in  the  case  of  each  of  the  N  original  per- 
mutations, the  whole  number  of  permutations  would  be  iVx  [p. 


450  ALGEBRA 

Again,  if  in  any  one  of  the  latter  the  q  6's  were  replaced  by 
q  new  letters,  differing  from  each  other  and  from  the  remain- 
ing letters,  then  by  altering  the  order  of  these  q  letters  among 
themselves,  we  could  from  the  original  permutation  form  [g 
different  permutations ;  and  if  this  were  done  in  the  case  of 
each  of  the  Nx\p_  permutations,  the  whole  number  of  permu- 
tations would  be  Nx\^X\q- 

In  like  manner,  if  in  each  of  the  latter  the  r  c's  were  replaced 
by  r  new  letters,  differing  from  each  other  and  from  the  remain- 
ing letters,  and  these  r  letters  were  permuted  among  them- 
selves, the  whole  number  of  permutations  would  be 

We  now  have  the  original  n  letters  replaced  by  n  different 
letters. 

But  the  number  of  permutations  of  n  different  things  taken 

n  at  a  time  is  [n  (§  501). 

In 
Therefore,  Nx\2JX\qx\r  =  \n:  or,  N=  ,    ,    .    • 

\p\q\L 

Any  other  case  can  be  treated  in  a  similar  manner. 

Ex.  How  many  permutations  can  be  formed  from  the  let- 
ters in  the  word  Tennessee,  taken  all  together  ? 

Here  there  are  4  e's,  2  n's,  2  s's,  and  1  t. 

Putting  in  the  above  formula  w  =  9,  jp  =  4,  g  =  2,  r  =  2,  we  have 

[9         5.6.7.8.9 


[4[2[2  2-2 


=  3780. 


EXERCISE  205 

1.  In  how  many  different  orders  can  the  letters  of  the  word 
denomination  be  written  ? 

2.  There  are  4  white  billiard  balls  exactly  alike,  and  3  red  balls, 
also  alike ;  in  how  many  different  orders  can  they  be  arranged  ? 

3.  In  how  many  different  orders  can  the  letters  of  the  word 
independence  be  written  ? 


PERMUTATIONS   AND   COMBINATIONS  451 

4.  How  many  different  signals  can  be  made  with  7  flags,  of 
which  2  are  blue,  3  red,  and  2  white,  if  all  are  hoisted  for  each 
signal  ? 

5.  How  many  different  numbers  of  8  digits  can  be  formed 
from  the  digits  4,  4,  3,  3,  3,  2,  2,  1  ? 

6.  In  how  many  different  ways  can  2  dimes,  3  quarters, 
4  halves,  and  5  dollars  be  distributed  among  14  persons,  so 
that  each  may  receive  a  coin? 

507.  To  find  for  what  value  of  r  the  number  of  combinations 
ofn  different  things  taken  r  at  a  time  is  greatest. 

By  §  502,  the  number  of  combinations  of  n  different  things^ 
taken  r  at  a  time,  is 

"^^-  1.2.3...  (r-l)r  ^^ 

Also,  the  number  of  combinations  of  n  different  things,  taken 
r  —  1  at  a  time,  is 

y,(,,_l)...[^_(^-l)  +  l]  n(n-l)  ...(71-^  +  2)      ,2>> 

1.2.3... (r-1)  '  1.2.3... (r-1)  ^^ 

The  expression  (1)  is  obtained  by  multiplying  the  expres- 

sion  (2)  by  — ,  or  — ' 1. 

r  r 

The  latter  expression  decreases  as  r  increases. 
If,  then,  we  find  the  values  of  (1)  corresponding  to  the  val- 
ues 1,  2,  3,  •  • .,  of  /•,  the  results  will  continually  increase  so 

long  as  ■ —  IS  >  1. 

r 

I.  Suppose  n  even ;  and  let  w  =  2  m,  where  m  is  a  positive 
integer. 

Then,  "-'•  +  ^  becomes  ?JItszI+l. 
r  r 

If  r  =  m,     ^~^"^     becomes  ^^        ,  and  is  >1. 
r  771 

If  r  =  m  + 1,  — — ^tL_  becomes  — — — ,  and  is  <  1. 

r  m  +  1 


452  ALGEBRA 


Then,  ^C^  will  have  its  greatest  value  when  r=m  —  -' 

II.    Suppose  n  odd ;  and  let  n  —  2m-\-l,  where  m  is  a  posi- 
tive integer. 

Then,  ^1=1+1  becomes  2«^^r±2. 
r  r 

If  r  =  m,     ^~  ^  "*"     becomes  ^"*"    ,  and  is  >1. 
r  m 

If  r  =  m  + 1,  -^  ^^  ~  ^  +  ^  becomes  ^  "^    ,  and  equals  1. 
r  m  +  1 

If  r  =  m  +  2,  ^"^-^^'^^  becomes  -^^,  and  is  <  1. 
r  m  +  2 

Then,   „C^   will  have  its   greatest  value  when  r  equals  m 

or  m  + 1 ;  that  is,  ^'^—^  or  ^^^  +  1. 

1 

Then,  „(7,.  will  have  its  greatest  value  when  r  equals  ^  ~ 

n4-l  ^ 

or      I'    ;  the  results  being  the  same  in  these  two  cases. 


EXPONENTIAL  AND  LOGARITHMIC   SERIES      453 


XLI.    EXPONENTIAL  AND  LOGARITHMIC 
SERIES 

508.  The  Theorem  of  Limits. 

If  two  expressions,  containing  the  same  variable  (§  317),  are 
equal  for  every  value  of  the  variable,  and  each  approaches  a 
limit  (§  318),  the  limits  are  equal. 

Let  A  and  B  be  two  expressions  containing  the  same  variable. 

Let  A  and  B  be  equal  for  every  value  of  the  variable,  and 
approach  the  limits  A'  and  B\  respectively. 

To  prove  A'=B\ 

Let  ^'— ^  =  m,  and  5'— JB  =  w. 

Then,  m  and  n  are  variables  which  can  be  made  less  than 
any  assigned  fixed  number,  however  small  (§  318). 

Then,  either  m  —  n  is  a  variable  which  can  be  made  less  than 
any  assigned  fixed  number,  however  small,  or  else  m  —  n  =  0. 

But  m  -  w  =  A-  A  -  (B'-  B) 

=  A'-A-B'-{-B  =  A'-B'; 

for,  by  hypothesis,  A  and  B  are  equal  for  every  value  of  the 
variable. 

But  A'—B'  is  not  a  variable;  and  hence  m  —  n  is  not  a 
variable. 

Then,  m  —  n  is  0;    and  hence  its  equal,  A'—B',  is  0,  or 

A'=B'. 

THE  EXPONENTIAL  SERIES 

509.  We  have  for  all  values  of  w  and  x. 


1+^ 

n 


=".!)■ 


454  ALGEBRA 

Expanding  both  members  by  the  Binomial  Theorem, 

[_  n  \2  n^  [3  n^         j 

^    ,  1  ,  nx(nx  —  l)     1 

n  [2  n^ 

^  |3  n«^     *      ^  ^ 

We  may  write  equation  (1)  in  the  form 


a;(  a; )      x( x Ycc 

=  ^  +  -  +  V^+^       f       "-  +  -S        (2) 

which  holds  however  great  n  may  be. 
Now  let  n  be  indefinitely  increased. 

1    2 
Then,  the  limit  of  each  of  the  terms  -,  -,  etc.,  is  0  (§  320). 

n    n 

Hence,  the  limiting  value  of  the  first  member  of  (2)  is 


['+'+tt-} 


and  the  limiting  value  of  the  second  member  is 

By  the  Theorem  of  Limits  (§  508),  these  limits  are  equal ; 
that  is. 

Denoting  the  series  in  brackets  by  e,  we  obtain 


EXPONENTIAL  AND  LOGARITHMIC   SERIES      455 

510.  Putting  mx  for  x,  in  (3),  §  509,  we  have 

e-  =  H-mx  +  ^  +  ^+....  (4) 

Let  m  =  logg  a. 

Then,  by  §  412,  e""  =  a,  and  e'"==  =  a*. 

Substituting  in  (4),  we  "obtain 

a^  =  1  +  (log,  a)x-^  (log,  ay^  +  (log,  af^  +  •  •  • .  (5) 

This  result  is  called  the  Exponential  Series. 

511.  The  system  of  logarithms  which  has  e  for  its  base 
is  called  the  Napierian  System,  from  Napier,  the  inventor  of 
logarithms. 

Napierian  logarithms  are  also  called  Natural  Logarithms. 
The  approximate  value  of  e  may  be  readily  calculated  from 
the  series  of  §  509. 

and  will  be  found  to  equal  2.7182818  ••.. 


THE  LOGARITHMIC  SERIES 

512.    To  expand  log,  (1  +  x)  in  ascendijig  powers  of  x. 
Substituting  in  (5),  §  510,  1  +  a;  for  a,  and  y  for  x, 

(1 H-  a;)^  =  1  +  [loge  (1  +  ic)]  ?/  H-  terms  in  /,  if,  etc. 
Expanding  the  first  member  by  the  Binomial  Theorem, 

=  1  +  [log, (1  +  a;)]  y  +  terms  in  y\  y^,  etc.  (6) 

This  equation  holds  for  every  value  of  y  which  makes  both 
members  convergent;  and,  by  the  Theorem  of  Undetermined" 
Coefficients  (§  396),  the  coefficients  of  y  in  the  two  series  are 
equal.  ^ 


456  ALGEBRA 

That  is,     x-^x^-{-^a^-^^x'-\-  ...  =  \og,(l^x), 

|Z  [^  [4 

Or,  log,(l  +  aj)  =  a.-|  +  J-|  +  |-....  (7) 

This  result  is  called  the  Logarithmic  Series. 


CALCULATION  OF  LOGARITHMS 

513.  The  equation  (7),  §  512,  can  be  used  to  calculate 
Napierian  Logarithms,  if  x  is  so  taken  that  the  second  mem- 
ber is  convergent;  but  unless  x  is  small,  it  requires  the  sum 
of  a  great  many  terms  to  insure  any  degree  of  accuracy. 

We  will  now  derive  a  more  convenient  series  for  the  calcula- 
tion of  Napierian  Logarithms. 


Bev         ;  2      3      4      5  ^^ 


514.  Putting  —  X  for  x,  in  (7),  §  512,  we  have 

log,(l-a;)  =  -a;----- 

Subtracting  (8)  from  (7),  we  obtain 

log.(l  +  a^)-logXl-a;)  =  2a^  +  ?|!  +  ?|!+.... 

o         o 

Or  (§422),  log.l±|  =  2(x  +  |  +  J  +  ...).  (9) 

^      m  —  n 

Let       x  = :  then  ~ —  = ! —  =  ---  =  — 

m  +  7i  1  —  x  m  —  n      2n      n 

m-\-n 
Substituting  these  values  in  (9),  we  obtain 


m  —  n     1  /m  —  nV     1  /m  —  nY  1 

m  +  n     3\m-\-nJ      5\m-\-nj  J 


m 


,    But  by  §  422,  log,— =  log,  m  —  log,  n ;  whence, 


n 


log,m  =  log,»  +  2r'5^  +  lf2^ 

|_m  +  71      3\m-\- 


n  Y  I  1  /m  —  nY      ^ 
n)       5lm  + 


EXPONENTIAL   AND   LOGARITHMIC   SERIES      457 

515.   Let   it   be    required,    for   example,    to   calculate   the 
Napierian  logarithm  of  2  to  six  places  of  decimals. 
Putting  m  =  2  and  n  =  l  in  the  result  of  §  514,  we  have 


log,2  =  log,l  +  2 


1  .  iriy  .  1/r 


3~^3VSj  '^BKSJ  "^ 


••} 


Or  since  log,l  =  0  (§418), 

loge  2  =  2(.3333333  +  .0123457  +  .0008230  +  .0000653 
+  .0000056  +  .0000005  +  •  •  •) 
=  2  X  .3465734  =  .6931468  =  .693147, 

correct  to  six  places  of  decimals. 

Having  found  log^2,  we  may  calculate  loge3  by  putting  m=3 
and  n  =  2  in  the  result  of  §  514. 

Proceeding  in  this  way,  we  shall  find  log^  10  =  2.302585  •••. 

516.  To  calculate  the  common  logarithm  of  a  number,  having 
given  its  Napieriayi  logarithm. 

Putting  6  =  10  and  a  =  e  in  the  result  of  §  426, 

Thus,  logio  2  =  .4342945  x  .693147  =  .301030. 

The  multiplier  by  .which  logarithms  of  any  system  are 
derived  from  Napierian  logarithms  is  called  the  modulus  of 
that  system. 

Thus,  .4342945  is  the  modulus  of  the  common  system. 

Conversely,  to  find  the  Napierian  logarithm  of  a  number 
when  its  common  logarithm  is  given,  we  may  either  divide  the 
common  logarithm  by  the  modulus  .4342945,  or  multiply  it  by 
2.302585,  the  reciprocal  of  .4342945. 

EXERCISE  206 

Using  the  table  of  common  logarithms,  find  the  Napierian 
logarithm  of  each  of  the  following  to  four  significant  figures : 

1.   10000.  2.   .001.  3.  9.93. 


458  *  ALGEBRA 

4.  243.6.  5.   .04568.  6.   .56734. 

7.  What  is  the  characteristic  of  logs  "^8  ? 

8.  What  is  the  characteristic  of  logj500? 

9.  If  log3  =  .4771,  how  many  digits  are  there  in  3^^  ? 
10.  If  log  8  =  .9031,  how  many  digits  are  there  in  8^^? 


INDEX 


Addition,  of  fractions,  109. 

of  imaginary  numbers,  243. 

of  monomials,  18. 

of  polynomials,  21. 

of    positive    and    negative    num- 
bers, 12. 

of  similar  terms,  19. 

of  surds,  226. 
Affected  quadratic  equations,  250. 
Any  power,  of  a  monomial,  63. 

of  a  fraction,  186, 
Any  root,  of  a  fraction,  192. 

of  a  monomial,  190. 
Approximate    square     root    of    an 

arithmetical  number,  201. 
Associative  Law,  for  addition,  410. 

for  multiplication,  411, 
Calculation  of  Logarithms,  456. 
Clearing  of  fractions,  53. 
Commutative  Law,  for  addition,  410. 

for  multiplication,  410. 
Completing  square,  first  method,  250. 

second  method,  253. 
Cube,  of  a  binomial,  188. 
Cube  root,  of  an  arithmetical  num- 
ber, 206. 

of  a  polynomial,  202. 
Definitions: 

Abscissa,  173. 

Absolute  Value,  12. 

Affected  Quadratic  Equation,  248. 

Algebraic  Expression,  9. 

Arithmetic  Means,  335. 

Arithmetic  Progression,  331. 

Arithmetical  Complement,  390. 

Axiom,  2. 

Binomial,  21. 

Characteristic,  377. 


Definitions  —  Continued 
Coefficient,  17. 
Combinations,  445. 
Common  Factor,  74. 
Common  Logarithm,  376. 
Common  Multiple,  100. 
Complex  Fraction,  121. 
Complex  Number,  242. 
Convergent  Series,  358. 
Cyclo-symmetric  ExpreaBion,  416. 
Degree  of  Equation,  52. 
Degree  of  Expression,  41. 
Divergent  Series,  358. 
Division,  42. 
Equation,  2. 

Equation  in  Quadratic  Form,  268. 
Equation  of  Condition,  51.  ""^ 
Equivalent  Equations,  146. 
Equivalent  Systems  of  Equations, 

431. 
Exponent,  7. 

Exponential  Equation,  393. 
Exponential  Series,  455. 
Factor,  17. 
Fraction,  103. 
Fractional  Exponent,  213. 
Geometric  Means,  344. 
Geometric  Progression,  338. 
Graph,  175.- 
Harmonic  Means,  347. 
Harmonic  Progression,  346. 
Homogeneous  Terms,  41. 
Identical  Equation,  51. 
Imaginary  Number,  242. 
Inconsistent  Equations,  147. 
Independent  Equations,  146. 
Indeterminate  Equations,  146. 
Index  of  Root,  190. 


459 


460 


INDEX 


Definitions  —  Continued 

Inequality,  180. 

Infinite  series,  350. 

Infinity,  305. 

Integral  Equation,  51. 

Irrational  Number,  222. 

Limit,  304. 

Linear  Equation,  52. 

Literal  Equation,  132. 

Mantissa,  377. 

Monomial,  17. 

Napierian  Logarithm,  455. 

Negative  Exponent,  214. 

Negative  Number,  12. 

Negative  Term,  17. 

Numerical  Equation,  51. 

Ordinate,  173. 

Perfect  Cube,  88. 

Perfect  Square,  76. 

Permutations,  445. 

Polynomial,  21. 

Positive  Number,  12. 

Positive  Term,  17. 

Proportion,  312. 

Pure  Quadratic  Equation,  248. 

Quadratic  Equation,  248. 

Quadratic  Expression,  276. 

Quadratic  Surd,  222. 

Rational  and  Integral,  40. 

Rational  Number,  222. 

Rectangular  Co-ordinates,  173. 

Root  of  Equation,  52. 

Series,  350. 

Similar  Surds,  226. 

Similar  Terms,  18. 

Simultaneous  Equations,  147." 

Subtraction,  18. 

Surd,  222. 

Symmetrical  Expression,  416. 

Trinomial,  21. 

Variable,  304. 

Zero  Exponent,  213. 
Discussion  of  general  quadratic  equa- 
tion, 281. 
Distributive    Law,  for    multiplica- 
tion, 412. 


Division,  by  detached  coefficients, 
441. 
of  fractions,  118. 
of  imaginary  numbers,  246. 
of  monomials,  43. 
of  polynomials  by  monomials,  45. 
of  polynomials  by  polynomials,  46. 
of  surds,  230. 
Elimination,  by  addition  or  subtrac- 
tion, 147. 
by  comparison,  150. 
by  substitution,  149. 
Evolution  of  surds,  233. 
Expansion,  of  fractions  into  series, 
361. 
of  surds  into  series,  363. 
Exponential  equations,  393. 
Extraction    of    roots  by  the   Bino- 
mial Theorem,  375. 
Factor  Theorem,  414. 
Factoring,     of     expressions    whose 
terms  have  a  common  factor, 
75,  76. 
of  quadratic  expressions,  276. 
of  symmetrical  expressions,  419. 
of  the  difference  of  two  perfect 

squares,  79. 
of  the  difference  of  any  two  equal 

odd  powers,  89. 
of  the  sum  or  difference  of  two 

perfect  cubes,  88, 
of  the  type  x*  +  ax'^y'^  +  yS  81,  279. 
of  the  type  x'^  -\- ax -\-  6,  82,  276. 
of  the  type  ax'^  +  6x  -h  c,  85,  276. 
of  trinomial  perfect  squares,  77. 
Formation  of  quadratic   equations, 

275. 
General  term  of    binomial   expan- 
sion, 355. 
Graphical  representation,   of   addi- 
tion of  complex  numbers,  436. 
of  complex  numbers,  435. 
of  imaginary  unit,  434. 
of  roots  of  equations,  179,  284. 
of  solutions  of  simultaneous  linear 
equations,  176. 


INDEX 


461 


of  solutions  of  simultaneous  quad- 
ratic equations,  301. 
Graph,  of  first  member  of  a  quad- 
ratic equation  having  equal  or 
imaginary  roots,  284. 

of  inconsistent  linear  equations 
with  two  unknown  numbers, 
177. 

of  indeterminate  linear  equations 
with  two  unknown  numbers, 
178. 

of  a  linear  equation  with  two  un- 
known numbers,  174. 

of  a  linear  expression  involving 
one  unknown  number,  178. 

of  a  quadratic  equation  involving 
two  unknown  numbers,  300. 

of  a  quadratic  expression  involv- 
ing one  unknown  number,  283. 
Graphs  in  Physics,  327. 
Highest  Common  Factor,  of  expres- 
sions which  can  be  readily  fac- 
tored by  inspection,  98. 

by  long  division,  395. 

Indeterminate  form  - ,  438. 

Indeterminate        forms,      0  x  oo , 

CO  -  00  ,  439. 
Interpretation,  of  solutions,  171. 

of  the  form  ^,  304. 
of  the  form  -  ,  305. 

GO 

Introduction  of  the  coefficient  of  a 
surd  under  the  radical  sign,  225. 

Involution  of  surds,  231. 

Logarithm  of  a  number  to  any  base, 
394. 

Lowest  Common  Multiple,   of  ex- 
pressions which  can  be  readily 
factored  by  inspection,  100. 
by  long  division,  401. 

Meaning  of  a  pure  imaginary  num- 
ber, 243. 

Multiplication,  of  fractions,  115. 
of  imaginary  numbers,  244. 


of  monomials,  33. 

of  polynomials  by  monomials, 
34. 

of  polynomials  by  polynomials,  35. 

of  positive  and  negative  numbers, 
14. 

of  surds,  228. 
Parentheses,  insertion  of,  30. 

removal  of,  28. 
Partial  fractions,  364. 
Permutations  of  things  not  all  dif- 
ferent taken  all  together,  449. 
Physical   Problems,   141,  260,   266, 

297,  325. 
Problem  of  the  Couriers,  306. 
Product,  of  the  sum  and  difference 
of  two  numbers,  6Q. 

of  two  binomials  having  same  first 
term,  67. 
Proof,   of  a"*  X  a"  =  «"*+»',   for  all 
values  of  m  and  7i,  405. 

of  Binomial  Theorem,  for  a  posi- 
tive integral  exponent,  350. 
Quadratic  surds,  237. 
Reduction,  of  fractions  to  integral 
or  mixed  expressions,  106. 

of  fractions  to  their  lowest  com- 
mon denominator,  107. 

of  fractions  to  their  lowest  terms 
when  the  numerator  and  de- 
nominator can  be  readily  fac- 
tored by  inspection,  104. 

of  fractions  to  their  lowest  terms 

•  when  the  numerator  and  de- 
nominator cannot  be  readily 
factored  by  inspection,  404, 

of  fractions  with  irrational  de- 
nominators to  equivalent  frac- 
tions with  rational  denomina- 
tors, 233. 

of  fractions  with  irrational  de- 
nominators to  equivalent  frac- 
tions with  rational  denomina- 
tors, when  the  denominators 
are  in  the  forms  a  ±  V&, 
Va  ±  ^^b,    or  7a  ±  V&,  405. 


462 


INDEX 


Reduction,  of  mixed  expressions  to 
fractions,  114. 
of  surds    of  different   degrees  to 
equivalent   surds   of   the  same 
degree,  227. 
of  surds  to  their  simplest  forms, 
222. 
Remainder  Theorem,  413. 
Repeating  decimals,  343. 
Reversion  of  series,  370. 
Solution,  of  equations  by  factoring, 
94,  280. 
of  equations  having  the  unknown 
numbers    under   radical  signs, 
239. 
of  equations  involving   decimals, 

134. 
of    fractional     linear    equations, 

127. 
of     integral     linear     equations, 
54. 

of  literal  affected  quadratic  equa- 
tions, 258. 


of  literal  linear  equations,  132. 
of  quadratic  equations   by   form- 
ula, 255. 
Square,  of  a  binomial,  64. 

of  a  polynomial,  186. 
Square    root,    of     an     arithmetical 
number,  197. 
of  a  binomial  surd,  238. 
of  a  polynomial,  193. 
Subtraction,  of  fractions,  109. 
of  imaginary  numbers,  243. 
of  monomials,  24. 
of  polynomials,  26. 
of  surds,  226. 
Sum  of  a  geometric  progression  to 

infinity,  342. 
Sum  and  product  of  roots  of  quad- 
ratic equations,  273. 
Theorem  of  Limits,  453. 
Theorem    of   Undetermined   Coeffi- 
cients (Rigorous),  409. 
Transposing  terms,  53. 
Use  of  Table  of  Logarithms,  383. 


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